এ অধ্যায়ে আমরা যে বিষয় গুলি আলোচনা করব।
- সার সংক্ষেপ
- উপ-গুণিতক কোণের ত্রিকোণমিতিক অনুপাত
- \(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
- \(1+\sin{A}=(\sin{\frac{A}{2}}+\cos{\frac{A}{2}})^2\)
- \(1-\sin{A}=(\sin{\frac{A}{2}}-\cos{\frac{A}{2}})^2\)
- \(\sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
- \(\cos{A}=\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}\)
- \(\cos{A}=2\cos^2{\frac{A}{2}}-1\)
- \(\cos{A}=1-2\sin^2{\frac{A}{2}}\)
- \(1+\cos{A}=2\cos^2{\frac{A}{2}}\)
- \(1-\cos{A}=2\sin^2{\frac{A}{2}}\)
- \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
- \(\frac{1-\cos{A}}{1+\cos{A}}=\tan^2{\frac{A}{2}}\)
- \(\tan{A}=\frac{2\tan{\frac{A}{2}}}{1-\tan^2{\frac{A}{2}}}\)
- \(\cot{A}=\frac{\cot^2{\frac{A}{2}}-1}{2\cot{\frac{A}{2}}}\)
- \(\sin{A}=3\sin{\frac{A}{3}}-4\sin^3{\frac{A}{3}}\)
- \(\cos{A}=4\cos^3{\frac{A}{3}}-3\cos{\frac{A}{3}}\)
- \(\tan{A}=\frac{3\tan{\frac{A}{3}}-\tan^3{\frac{A}{3}}}{1-3\tan^2{\frac{A}{3}}}\)
- \(\cot{A}=\frac{\cot^3{\frac{A}{3}}-3\cot{\frac{A}{3}}}{3\cot^2{\frac{A}{3}}-1}\)
- \(\cot{\frac{A}{2}}-\tan{\frac{A}{2}}=2\cot{A}\)
- \(15^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- \(18^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- \(36^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- \(54^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- \(72^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- \(75^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
- অধ্যায় \(7E\)-এর উদাহরণসমুহ
- অধ্যায় \(7E\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(7E\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(7E\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(7E\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
সার সংক্ষেপ
উপ-গুণিতক কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\) \(1+\sin{A}=(\sin{\frac{A}{2}}+\cos{\frac{A}{2}})^2\) \(1-\sin{A}=(\sin{\frac{A}{2}}-\cos{\frac{A}{2}})^2\) \(\sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\) \(\cos{A}=\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}\) \(\cos{A}=2\cos^2{\frac{A}{2}}-1\) \(\cos{A}=1-2\sin^2{\frac{A}{2}}\) \(1+\cos{A}=2\cos^2{\frac{A}{2}}\) \(1-\cos{A}=2\sin^2{\frac{A}{2}}\) \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\) \(\frac{1-\cos{A}}{1+\cos{A}}=\tan^2{\frac{A}{2}}\) \(\tan{A}=\frac{2\tan{\frac{A}{2}}}{1-\tan^2{\frac{A}{2}}}\) \(\cot{A}=\frac{\cot^2{\frac{A}{2}}-1}{2\cot{\frac{A}{2}}}\) \(\sin{A}=3\sin{\frac{A}{3}}-4\sin^3{\frac{A}{3}}\) \(\cos{A}=4\cos^3{\frac{A}{3}}-3\cos{\frac{A}{3}}\) \(\tan{A}=\frac{3\tan{\frac{A}{3}}-\tan^3{\frac{A}{3}}}{1-3\tan^2{\frac{A}{3}}}\)\(\cot{A}=\frac{\cot^3{\frac{A}{3}}-3\cot{\frac{A}{3}}}{3\cot^2{\frac{A}{3}}-1}\)\(\cot{\frac{A}{2}}-\tan{\frac{A}{2}}=2\cot{A}\)\(15^{o}\) কোণের ত্রিকোণমিতিক অনুপাত \(18^{o}\) কোণের ত্রিকোণমিতিক অনুপাত \(36^{o}\) কোণের ত্রিকোণমিতিক অনুপাত \(54^{o}\) কোণের ত্রিকোণমিতিক অনুপাত \(72^{o}\) কোণের ত্রিকোণমিতিক অনুপাত \(75^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
উপ-গুণিতক কোণের ত্রিকোণমিতিক অনুপাত
Trigonometric ratio of sub-multiple angles
একটি কোণকে কোনো পূর্ণসংখ্যা দ্বারা ভাগ করলে উক্ত কোণের উপ-গুণিতক কোণ পাওয়া যায়।
যেমনঃ \(A\) কোণের উপ-গুণিতক কোণগুলি \(\frac{A}{2}, \ \frac{A}{3}, \ \frac{A}{4} .........\frac{A}{n}\) ইত্যাদি।
যেমনঃ \(A\) কোণের উপ-গুণিতক কোণগুলি \(\frac{A}{2}, \ \frac{A}{3}, \ \frac{A}{4} .........\frac{A}{n}\) ইত্যাদি।
প্রয়োজনীয় ও স্বরণীয় সূত্রসমূহ
Necessary and memorable formulas
\(\sin{A}\) কে \(\sin{\frac{A}{2}}\) এবং \(\cos{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\) \(1+\sin{A}=(\sin{\frac{A}{2}}+\cos{\frac{A}{2}})^2\) \(1-\sin{2}=(\sin{\frac{A}{2}}-\cos{\frac{A}{2}})^2\) \(\sin{A}\) কে \(\tan{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\) \(\cos{A}\) কে \(\sin{\frac{A}{2}}\) এবং \(\cos{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\cos{A}=\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}\) \(\cos{A}=2\cos^2{\frac{A}{2}}-1\) \(\cos{A}=1-2\sin^2{\frac{A}{2}}\) \(1+\cos{A}=2\cos^2{\frac{A}{2}}\) \(1-\cos{A}=2\sin^2{\frac{A}{2}}\) \(\cos{A}\) কে \(\tan{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\) \(\frac{1-\cos{A}}{1+\cos{A}}=\tan^2{\frac{A}{2}}\) \(\tan{A}\) কে \(\tan{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\tan{A}=\frac{2\tan{\frac{A}{2}}}{1-\tan^2{\frac{A}{2}}}\) \(\cot{A}\) কে \(\cot{\frac{A}{2}}\) এর মাধ্যমে প্রকাশ
\(\cot{A}=\frac{\cot^2{\frac{A}{2}}-1}{2\cot{\frac{A}{2}}}\) \(\sin{A}\) কে \(\sin{\frac{A}{3}}\) এর মাধ্যমে প্রকাশ
\(\sin{A}=3\sin{\frac{A}{3}}-4\sin^3{\frac{A}{3}}\) \(\cos{A}\) কে \(\cos{\frac{A}{3}}\) এর মাধ্যমে প্রকাশ
\(\cos{A}=4\cos^3{\frac{A}{3}}-3\cos{\frac{A}{3}}\) \(\tan{A}\) কে \(\tan{\frac{A}{3}}\) এর মাধ্যমে প্রকাশ
\(\tan{A}=\frac{3\tan{\frac{A}{3}}-\tan^3{\frac{A}{3}}}{1-3\tan^2{\frac{A}{3}}}\) \(\cot{A}\) কে \(\cot{\frac{A}{3}}\) এর মাধ্যমে প্রকাশ
\(\cot{A}=\frac{\cot^3{\frac{A}{3}}-3\cot{\frac{A}{3}}}{3\cot^2{\frac{A}{3}}-1}\) \(15^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{15^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\) \(\cos{15^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\) \(\tan{15^{o}}=2-\sqrt{3}\)\(cosec \ {15^{o}}=\sqrt{6}+\sqrt{2}\) \(\sec{15^{o}}=\sqrt{6}-\sqrt{2}\) \(\cot{15^{o}}=2+\sqrt{3}\)
\(18^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\) \(\tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)\(cosec \ {18^{o}}=\sqrt{5}+1\) \(\sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\) \(\cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
\(36^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\) \(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\) \(\tan{36^{o}}=\sqrt{5-2\sqrt{5}}\)\(cosec \ {36^{o}}=\frac{1}{5}\sqrt{50+10\sqrt{5}}\) \(\sec{36^{o}}=\sqrt{5}-1\) \(\cot{36^{o}}=\frac{1}{5}\sqrt{25+10\sqrt{5}}\)
\(54^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{54^{o}}=\frac{1}{4}(\sqrt{5}+1)\) \(\cos{54^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\) \(\tan{54^{o}}=\frac{1}{5}\sqrt{25+10\sqrt{5}}\)\(cosec \ {54^{o}}=\sqrt{5}-1\) \(\sec{54^{o}}=\frac{1}{5}\sqrt{50+10\sqrt{5}}\) \(\cot{54^{o}}=\sqrt{5-2\sqrt{5}}\)
\(72^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{72^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\) \(\cos{72^{o}}=\frac{1}{4}(\sqrt{5}-1)\) \(\tan{72^{o}}=\sqrt{5+2\sqrt{5}}\)\(cosec \ {72^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\) \(\sec{72^{o}}=\sqrt{5}+1\) \(\cot{72^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(75^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{75^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\) \(\cos{75^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\) \(\tan{75^{o}}=2+\sqrt{3}\)\(cosec \ {75^{o}}=\sqrt{6}-\sqrt{2}\) \(\sec{75^{o}}=\sqrt{6}+\sqrt{2}\) \(\cot{75^{o}}=2-\sqrt{3}\)
দ্রষ্টব্যঃ
\(15^{o}, \ 18^{o}, \ 30^{o}, \ 36^{o}\) ও \(45^{o}\) কোণের ত্রিকোণমিতিক অনুপাত জানা থাকলে নিম্নলিখিত কোণের ত্রিকোণমিতিক অনুপাত নির্ণয় করা যায়। \(3^{o}=18^{o}-15^{o}, \ 6^{o}=36^{o}-30^{o}, \ 9^{o}=45^{o}-36^{o}, \ 12^{o}=30^{o}-18^{o}, \ 21^{o}=36^{o}-15^{o}\) ইত্যাদি।
\(18^{o}\) কোণের ত্রিকোণমিতিক অনুপাত
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\) \(\tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)\(cosec \ {18^{o}}=\sqrt{5}+1\) \(\sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\) \(\cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
প্রমাণঃ
ধরি,
\(\theta=18^{o}\)
\(\Rightarrow 5\theta=5\times18^{o}\)
\(\Rightarrow 5\theta=90^{o}\)
\(\Rightarrow 2\theta+3\theta=90^{o}\)
\(\Rightarrow 2\theta=90^{o}-3\theta\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}\times1-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\cos{3\theta}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow 2\sin{\theta}\cos{\theta}=4\cos^3{\theta}-3\cos{\theta}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{\theta}(4\cos^2{\theta}-3)\)
\(\Rightarrow 2\sin{\theta}=4\cos^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=4(1-\sin^2{\theta})-3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2\sin{\theta}=4-4\sin^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=1-4\sin^2{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+2\sin{\theta}-1=0\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{2^2-4\times4\times(-1)}}}{2\times4}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4+16}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{20}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4\times5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{2\sqrt{5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{2(-1\pm{\sqrt{5}})}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-1\pm{\sqrt{5}}}{4}\)
\(\Rightarrow \sin{18^{o}}=\frac{-1+\sqrt{5}}{4}\) ➜ \(\because 18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
তাই \(\sin{18^{o}}\) এর মান ধনাত্মক হবে।
\(\therefore \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
আবার,
\(\cos{18^{o}}=\sqrt{1-\sin^2{18^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}(\sqrt{5}-1)\right\}^2}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\sqrt{1-\frac{1}{16}(\sqrt{5}-1)^2}\)
\(=\sqrt{\frac{1}{16}\{16-(\sqrt{5}-1)^2\}}\)
\(=\frac{1}{4}\sqrt{16-(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{4}\sqrt{16-(6-2\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{16-6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\therefore \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
আবার,
\(\tan{18^{o}}=\frac{\sin{18^{o}}}{\cos{18^{o}}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\frac{\frac{1}{4}(\sqrt{5}-1)}{\frac{1}{4}\sqrt{10+2\sqrt{5}}}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
এবং \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\)
\(=\frac{(\sqrt{5}-1)\sqrt{10-2\sqrt{5}}}{\sqrt{10+2\sqrt{5}}\times\sqrt{10-2\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{10-2\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{(\sqrt{5}-1)\sqrt{2\sqrt{5}(\sqrt{5}-1)}}{\sqrt{100-20}}\) ➜ \(\because \sqrt{10-2\sqrt{5}}=\sqrt{2\sqrt{5}(\sqrt{5}-1)}\)
এবং \((a+b)(a-b)=a^2-b^2\)
\(=\frac{\sqrt{2\sqrt{5}(\sqrt{5}-1)(\sqrt{5}-1)^2}}{\sqrt{80}}\)
\(=\frac{\sqrt{2\sqrt{5}(\sqrt{5}-1)(5+1-2\sqrt{5})}}{4\sqrt{5}}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
এবং \(\sqrt{80}=4\sqrt{5}\)
\(=\frac{\sqrt{4\sqrt{5}(\sqrt{5}-1)(3-\sqrt{5})}}{4\sqrt{5}}\)
\(=\frac{2\sqrt{\sqrt{5}(3\sqrt{5}-3-5+\sqrt{5})}}{4\sqrt{5}}\)
\(=\frac{\sqrt{\sqrt{5}(4\sqrt{5}-8)}}{2\sqrt{5}}\)
\(=\frac{\sqrt{4\sqrt{5}(\sqrt{5}-2)}}{2\sqrt{5}}\)
\(=\frac{2\sqrt{\sqrt{5}(\sqrt{5}-2)}}{2\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{(5-2\sqrt{5})}}{\sqrt{5}\times\sqrt{5}}\) ➜ লব ও হরকে \(\sqrt{5}\) দ্বারা গুণ করে।
\(=\frac{\sqrt{25-10\sqrt{5}}}{5}\)
\(=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(\therefore \tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(\tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
আবার,
\(cosec \ {18^{o}}=\frac{1}{\sin{18^{o}}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
\(=\frac{1}{\frac{1}{4}(\sqrt{5}-1)}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\frac{4}{\sqrt{5}-1}\)
\(=\frac{4(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}\) ➜ লব ও হরকে \((\sqrt{5}+1)\) দ্বারা গুণ করে।
\(=\frac{4(\sqrt{5}+1)}{5-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{4(\sqrt{5}+1)}{4}\)
\(=\sqrt{5}+1\)
\(\therefore cosec \ {18^{o}}=\sqrt{5}+1\)
\(cosec \ {18^{o}}=\sqrt{5}+1\)
আবার,
\(\sec{18^{o}}=\frac{1}{\cos{18^{o}}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(=\frac{1}{\frac{1}{4}\sqrt{10+2\sqrt{5}}}\) ➜ \(\because \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{4}{\sqrt{10+2\sqrt{5}}}\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{10+2\sqrt{5}}\times\sqrt{10-2\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{10-2\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{100-20}}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{80}}\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{4\sqrt{5}}\)➜ \(\because \sqrt{80}=4\sqrt{5}\)
\(=\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{10-2\sqrt{5}}}{\sqrt{5}\times\sqrt{5}}\) ➜ লব ও হরকে \(\sqrt{5}\) দ্বারা গুণ করে।
\(=\frac{\sqrt{5(10-2\sqrt{5})}}{5}\)
\(=\frac{\sqrt{50-10\sqrt{5}}}{5}\)
\(=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
\(\therefore \sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
\(\sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
আবার,
\(\cot{18^{o}}=\frac{1}{\tan{18^{o}}}\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(=\frac{1}{\frac{1}{5}\sqrt{25-10\sqrt{5}}}\) ➜ \(\because \tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(=\frac{5}{\sqrt{25-10\sqrt{5}}}\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{25+10\sqrt{5}}\times\sqrt{25-10\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{25+10\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{625-500}}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{125}}\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{5\sqrt{5}}\)➜ \(\because \sqrt{125}=5\sqrt{5}\)
\(=\frac{\sqrt{25+10\sqrt{5}}}{\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{5+2\sqrt{5}}}{\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}}\)
\(\therefore \cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
\(\cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
\(\theta=18^{o}\)
\(\Rightarrow 5\theta=5\times18^{o}\)
\(\Rightarrow 5\theta=90^{o}\)
\(\Rightarrow 2\theta+3\theta=90^{o}\)
\(\Rightarrow 2\theta=90^{o}-3\theta\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}\times1-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\cos{3\theta}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow 2\sin{\theta}\cos{\theta}=4\cos^3{\theta}-3\cos{\theta}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{\theta}(4\cos^2{\theta}-3)\)
\(\Rightarrow 2\sin{\theta}=4\cos^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=4(1-\sin^2{\theta})-3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2\sin{\theta}=4-4\sin^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=1-4\sin^2{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+2\sin{\theta}-1=0\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{2^2-4\times4\times(-1)}}}{2\times4}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4+16}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{20}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4\times5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{2\sqrt{5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{2(-1\pm{\sqrt{5}})}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-1\pm{\sqrt{5}}}{4}\)
\(\Rightarrow \sin{18^{o}}=\frac{-1+\sqrt{5}}{4}\) ➜ \(\because 18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
তাই \(\sin{18^{o}}\) এর মান ধনাত্মক হবে।
\(\therefore \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
আবার,
\(\cos{18^{o}}=\sqrt{1-\sin^2{18^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}(\sqrt{5}-1)\right\}^2}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\sqrt{1-\frac{1}{16}(\sqrt{5}-1)^2}\)
\(=\sqrt{\frac{1}{16}\{16-(\sqrt{5}-1)^2\}}\)
\(=\frac{1}{4}\sqrt{16-(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{4}\sqrt{16-(6-2\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{16-6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\therefore \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
আবার,
\(\tan{18^{o}}=\frac{\sin{18^{o}}}{\cos{18^{o}}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\frac{\frac{1}{4}(\sqrt{5}-1)}{\frac{1}{4}\sqrt{10+2\sqrt{5}}}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
এবং \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\)
\(=\frac{(\sqrt{5}-1)\sqrt{10-2\sqrt{5}}}{\sqrt{10+2\sqrt{5}}\times\sqrt{10-2\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{10-2\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{(\sqrt{5}-1)\sqrt{2\sqrt{5}(\sqrt{5}-1)}}{\sqrt{100-20}}\) ➜ \(\because \sqrt{10-2\sqrt{5}}=\sqrt{2\sqrt{5}(\sqrt{5}-1)}\)
এবং \((a+b)(a-b)=a^2-b^2\)
\(=\frac{\sqrt{2\sqrt{5}(\sqrt{5}-1)(\sqrt{5}-1)^2}}{\sqrt{80}}\)
\(=\frac{\sqrt{2\sqrt{5}(\sqrt{5}-1)(5+1-2\sqrt{5})}}{4\sqrt{5}}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
এবং \(\sqrt{80}=4\sqrt{5}\)
\(=\frac{\sqrt{4\sqrt{5}(\sqrt{5}-1)(3-\sqrt{5})}}{4\sqrt{5}}\)
\(=\frac{2\sqrt{\sqrt{5}(3\sqrt{5}-3-5+\sqrt{5})}}{4\sqrt{5}}\)
\(=\frac{\sqrt{\sqrt{5}(4\sqrt{5}-8)}}{2\sqrt{5}}\)
\(=\frac{\sqrt{4\sqrt{5}(\sqrt{5}-2)}}{2\sqrt{5}}\)
\(=\frac{2\sqrt{\sqrt{5}(\sqrt{5}-2)}}{2\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{(5-2\sqrt{5})}}{\sqrt{5}\times\sqrt{5}}\) ➜ লব ও হরকে \(\sqrt{5}\) দ্বারা গুণ করে।
\(=\frac{\sqrt{25-10\sqrt{5}}}{5}\)
\(=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(\therefore \tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(\tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
আবার,
\(cosec \ {18^{o}}=\frac{1}{\sin{18^{o}}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
\(=\frac{1}{\frac{1}{4}(\sqrt{5}-1)}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\frac{4}{\sqrt{5}-1}\)
\(=\frac{4(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}\) ➜ লব ও হরকে \((\sqrt{5}+1)\) দ্বারা গুণ করে।
\(=\frac{4(\sqrt{5}+1)}{5-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{4(\sqrt{5}+1)}{4}\)
\(=\sqrt{5}+1\)
\(\therefore cosec \ {18^{o}}=\sqrt{5}+1\)
\(cosec \ {18^{o}}=\sqrt{5}+1\)
আবার,
\(\sec{18^{o}}=\frac{1}{\cos{18^{o}}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(=\frac{1}{\frac{1}{4}\sqrt{10+2\sqrt{5}}}\) ➜ \(\because \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{4}{\sqrt{10+2\sqrt{5}}}\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{10+2\sqrt{5}}\times\sqrt{10-2\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{10-2\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{100-20}}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{\sqrt{80}}\)
\(=\frac{4\sqrt{10-2\sqrt{5}}}{4\sqrt{5}}\)➜ \(\because \sqrt{80}=4\sqrt{5}\)
\(=\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{10-2\sqrt{5}}}{\sqrt{5}\times\sqrt{5}}\) ➜ লব ও হরকে \(\sqrt{5}\) দ্বারা গুণ করে।
\(=\frac{\sqrt{5(10-2\sqrt{5})}}{5}\)
\(=\frac{\sqrt{50-10\sqrt{5}}}{5}\)
\(=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
\(\therefore \sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
\(\sec{18^{o}}=\frac{1}{5}\sqrt{50-10\sqrt{5}}\)
আবার,
\(\cot{18^{o}}=\frac{1}{\tan{18^{o}}}\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(=\frac{1}{\frac{1}{5}\sqrt{25-10\sqrt{5}}}\) ➜ \(\because \tan{18^{o}}=\frac{1}{5}\sqrt{25-10\sqrt{5}}\)
\(=\frac{5}{\sqrt{25-10\sqrt{5}}}\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{25+10\sqrt{5}}\times\sqrt{25-10\sqrt{5}}}\) ➜ লব ও হরকে \(\sqrt{25+10\sqrt{5}}\) দ্বারা গুণ করে।
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{625-500}}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{\sqrt{125}}\)
\(=\frac{5\sqrt{25+10\sqrt{5}}}{5\sqrt{5}}\)➜ \(\because \sqrt{125}=5\sqrt{5}\)
\(=\frac{\sqrt{25+10\sqrt{5}}}{\sqrt{5}}\)
\(=\frac{\sqrt{5}\sqrt{5+2\sqrt{5}}}{\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}}\)
\(\therefore \cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
\(\cot{18^{o}}=\sqrt{5+2\sqrt{5}}\)
অধ্যায় \(7E\)-এর উদাহরণসমুহ
উদাহরণ \(1.\) প্রমাণ কর যে, \(\sin{\left(7\frac{1}{2}\right)^{o}}=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{3}}}\)
উদাহরণ \(2.\) প্রমাণ কর যে, \(\sin{\left(67\frac{1}{2}\right)^{o}}=\frac{1}{2}\sqrt{2+\sqrt{2}}\)
উদাহরণ \(3.\) প্রমাণ কর যে, \(2\sin{\left(\frac{\pi}{24}\right)}=\sqrt{2-\sqrt{2+\sqrt{3}}}\)
উদাহরণ \(4.\) প্রমাণ কর যে, \(2\sin{\left(\frac{\pi}{32}\right)}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\)
উদাহরণ \(5.\) প্রমাণ কর যে, \(2\sin{\frac{\pi}{16}}=2\sin{11^{o}15^{\prime}}=\sqrt{2-\sqrt{2+\sqrt{2}}}\)
উদাহরণ \(6.\) যদি \(\tan{\frac{\theta}{2}}=\sqrt{\frac{1-e}{1+e}}\tan{\frac{\phi}{2}}\) হয় তবে দেখাও যে, \(\cos{\phi}=\frac{\cos{\theta}-e}{1-e\cos{\theta}}\)
উদাহরণ \(7.\) \(\sin{\left(22\frac{1}{2}\right)^{o}}\) এবং \(\cos{\left(22\frac{1}{2}\right)^{o}}\) এর মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\sqrt{2-\sqrt{2}}, \ \frac{1}{2}\sqrt{2+\sqrt{2}}\)
উদাহরণ \(2.\) প্রমাণ কর যে, \(\sin{\left(67\frac{1}{2}\right)^{o}}=\frac{1}{2}\sqrt{2+\sqrt{2}}\)
উদাহরণ \(3.\) প্রমাণ কর যে, \(2\sin{\left(\frac{\pi}{24}\right)}=\sqrt{2-\sqrt{2+\sqrt{3}}}\)
উদাহরণ \(4.\) প্রমাণ কর যে, \(2\sin{\left(\frac{\pi}{32}\right)}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\)
উদাহরণ \(5.\) প্রমাণ কর যে, \(2\sin{\frac{\pi}{16}}=2\sin{11^{o}15^{\prime}}=\sqrt{2-\sqrt{2+\sqrt{2}}}\)
ঢাঃ ২০০৮;চঃ২০১৬,২০১৪; যঃ২০১৬,২০১৪,২০১০; দিঃ২০১২,২০১০; বঃ২০১৪,২০১০; রাঃ২০১৪,২০১২,২০০৫; কুঃ২০০৩; সিঃ ২০১৪; মাঃ ২০১১; বুটেক্সঃ২০০৭-৮; বুয়েটঃ২০১৩-২০১৪; চুয়েটঃ ২০০৫-২০০৬।
উদাহরণ \(6.\) যদি \(\tan{\frac{\theta}{2}}=\sqrt{\frac{1-e}{1+e}}\tan{\frac{\phi}{2}}\) হয় তবে দেখাও যে, \(\cos{\phi}=\frac{\cos{\theta}-e}{1-e\cos{\theta}}\)
ঢাঃ ২০১৪,২০০৬; রাঃ২০১৫,২০০৯; দিঃ২০১৬;কুঃ ২০১৫; চঃ ২০০৮; সিঃ২০১৫,২০১২,২০০৮; বঃ২০১৫; মাঃ ২০১৩; বুয়েটঃ২০০১-২০০২।
উদাহরণ \(7.\) \(\sin{\left(22\frac{1}{2}\right)^{o}}\) এবং \(\cos{\left(22\frac{1}{2}\right)^{o}}\) এর মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\sqrt{2-\sqrt{2}}, \ \frac{1}{2}\sqrt{2+\sqrt{2}}\)
উদাহরণ \(8.\) দেখাও যে,
\((a)\) \(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\) \(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\) \(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
উদাহরণ \(9.\) যদি \(\tan{\frac{\theta}{2}}=\tan^3{\frac{\phi}{2}}\) এবং \(\tan{\phi}=2tan{\alpha}\) হয় তবে দেখাও যে, \(\theta+\phi=2\alpha\)
উদাহরণ \(10.\) যদি \(\sin{\alpha}+\sin{\beta}=a\) এবং \(\cos{\alpha}+\cos{\beta}=b\) হয়, তবে দেখাও যে, \(\sin{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{4-a^2-b^2}\)
উদাহরণ \(11.\) দেখাও যে,
\((a)\) \(\sin{15^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\((b)\) \(\cos{15^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
\((c)\) \(\tan{15^{o}}=2-\sqrt{3}\)
উদাহরণ \(12.\) প্রমাণ কর যে, \(\sec{x}=\frac{2}{\sqrt{2+\sqrt{2+2\cos{4x}}}}\)
\((a)\) \(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\) \(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\) \(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
উদাহরণ \(9.\) যদি \(\tan{\frac{\theta}{2}}=\tan^3{\frac{\phi}{2}}\) এবং \(\tan{\phi}=2tan{\alpha}\) হয় তবে দেখাও যে, \(\theta+\phi=2\alpha\)
উদাহরণ \(10.\) যদি \(\sin{\alpha}+\sin{\beta}=a\) এবং \(\cos{\alpha}+\cos{\beta}=b\) হয়, তবে দেখাও যে, \(\sin{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{4-a^2-b^2}\)
বঃ, দিঃ২০১৭;কুঃ ২০১৯ ।
উদাহরণ \(11.\) দেখাও যে,
\((a)\) \(\sin{15^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\((b)\) \(\cos{15^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
\((c)\) \(\tan{15^{o}}=2-\sqrt{3}\)
উদাহরণ \(12.\) প্রমাণ কর যে, \(\sec{x}=\frac{2}{\sqrt{2+\sqrt{2+2\cos{4x}}}}\)
ঢাঃ২০১৪; দিঃ২০০৯; যঃ২০০৫।
উদাহরণ \(2.\) প্রমাণ কর যে, \(\sin{\left(67\frac{1}{2}\right)^{o}}=\frac{1}{2}\sqrt{2+\sqrt{2}}\)
সমাধানঃ
\(L.S=\sin{\left(67\frac{1}{2}\right)^{o}}\)
\(=\sin{\left(\frac{135}{2}\right)^{o}}\)
\(=\frac{1}{2}\sqrt{4\sin^2{\left(\frac{135}{2}\right)^{o}}}\)
\(=\frac{1}{2}\sqrt{2\times2\sin^2{\left(\frac{135}{2}\right)^{o}}}\)
\(=\frac{1}{2}\sqrt{2\left\{1-\cos{2\left(\frac{135}{2}\right)^{o}}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\sqrt{2(1-\cos{135^{o}})}\)
\(=\frac{1}{2}\sqrt{2-2\cos{(90^{o}\times2-45^{o})}}\) ➜ \(\because 135^{o}=90^{o}\times2-45^{o}\)
\(=\frac{1}{2}\sqrt{2+2\cos{45^{o}}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।
\(=\frac{1}{2}\sqrt{2+2\frac{1}{\sqrt{2}}}\) ➜ \(\because \cos{45^{o}}=\frac{1}{\sqrt{2}}\)
\(=\frac{1}{2}\sqrt{2+\frac{2}{\sqrt{2}}}\)
\(=\frac{1}{2}\sqrt{2+\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}}\)
\(=\frac{1}{2}\sqrt{2+\sqrt{2}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(=\sin{\left(\frac{135}{2}\right)^{o}}\)
\(=\frac{1}{2}\sqrt{4\sin^2{\left(\frac{135}{2}\right)^{o}}}\)
\(=\frac{1}{2}\sqrt{2\times2\sin^2{\left(\frac{135}{2}\right)^{o}}}\)
\(=\frac{1}{2}\sqrt{2\left\{1-\cos{2\left(\frac{135}{2}\right)^{o}}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\sqrt{2(1-\cos{135^{o}})}\)
\(=\frac{1}{2}\sqrt{2-2\cos{(90^{o}\times2-45^{o})}}\) ➜ \(\because 135^{o}=90^{o}\times2-45^{o}\)
\(=\frac{1}{2}\sqrt{2+2\cos{45^{o}}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।
\(=\frac{1}{2}\sqrt{2+2\frac{1}{\sqrt{2}}}\) ➜ \(\because \cos{45^{o}}=\frac{1}{\sqrt{2}}\)
\(=\frac{1}{2}\sqrt{2+\frac{2}{\sqrt{2}}}\)
\(=\frac{1}{2}\sqrt{2+\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}}\)
\(=\frac{1}{2}\sqrt{2+\sqrt{2}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
উদাহরণ \(8.\) দেখাও যে,
\((a)\) \(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\) \(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\) \(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\((a)\) \(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\) \(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\) \(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\) \(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
সমাধানঃ
\((a)\)
ধরি,
\(\theta=18^{o}\)
\(\Rightarrow 5\theta=5\times18^{o}\)
\(\Rightarrow 5\theta=90^{o}\)
\(\Rightarrow 2\theta+3\theta=90^{o}\)
\(\Rightarrow 2\theta=90^{o}-3\theta\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}\times1-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\cos{3\theta}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow 2\sin{\theta}\cos{\theta}=4\cos^3{\theta}-3\cos{\theta}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
এবং \(\cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{\theta}(4\cos^2{\theta}-3)\)
\(\Rightarrow 2\sin{\theta}=4\cos^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=4(1-\sin^2{\theta})-3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2\sin{\theta}=4-4\sin^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=1-4\sin^2{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+2\sin{\theta}-1=0\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{2^2-4\times4\times(-1)}}}{2\times4}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4+16}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{20}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4\times5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{2\sqrt{5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{2(-1\pm{\sqrt{5}})}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-1\pm{\sqrt{5}}}{4}\)
\(\Rightarrow \sin{18^{o}}=\frac{-1+\sqrt{5}}{4}\) ➜ \(\because 18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
তাই \(\sin{18^{o}}\) এর মান ধনাত্মক হবে।
\(\therefore \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\)
লেখা যায়,
\(\cos{36^{o}}=\sqrt{1-\sin^2{36^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(36^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}\sqrt{10-2\sqrt{5}}\right\}^2}\) ➜ \(\because \sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(=\sqrt{1-\frac{1}{16}(10-2\sqrt{5})}\)
\(=\sqrt{\frac{1}{16}\{16-(10-2\sqrt{5})\}}\)
\(=\frac{1}{4}\sqrt{16-10+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{5+1+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{(\sqrt{5}+1)^2}\) ➜ \(\because a^2+b^2+2ab=(a+b)^2\)
\(=\frac{1}{4}(\sqrt{5}+1)\)
\(\therefore \cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\)
লেখা যায়,
\(\cos{18^{o}}=\sqrt{1-\sin^2{18^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}(\sqrt{5}-1)\right\}^2}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\sqrt{1-\frac{1}{16}(\sqrt{5}-1)^2}\)
\(=\sqrt{\frac{1}{16}\{16-(\sqrt{5}-1)^2\}}\)
\(=\frac{1}{4}\sqrt{16-(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{4}\sqrt{16-(6-2\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{16-6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\therefore \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\)
লেখা যায়,
\(\sin{36^{o}}=\sin{(2\times18^{o})}\)
\(=2\sin{18^{o}}\cos{18^{o}}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=2\times\frac{1}{4}(\sqrt{5}-1)\times\frac{1}{4}\sqrt{10+2\sqrt{5}}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{1}{8}(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}\)
\(=\frac{1}{8}(\sqrt{5}-1)\sqrt{2\sqrt{5}(\sqrt{5}+1)}\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(\sqrt{5}-1)^2}\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(6-2\sqrt{5})}\)
\(=\frac{1}{8}\sqrt{4\sqrt{5}(\sqrt{5}+1)(3-\sqrt{5})}\)
\(=\frac{2}{8}\sqrt{\sqrt{5}(3\sqrt{5}+3-5-\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{\sqrt{5}(2\sqrt{5}-2)}\)
\(=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(\therefore \sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
ধরি,
\(\theta=18^{o}\)
\(\Rightarrow 5\theta=5\times18^{o}\)
\(\Rightarrow 5\theta=90^{o}\)
\(\Rightarrow 2\theta+3\theta=90^{o}\)
\(\Rightarrow 2\theta=90^{o}-3\theta\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\sin{(90^{o}\times1-3\theta)}\)
\(\Rightarrow \sin{2\theta}=\cos{3\theta}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow 2\sin{\theta}\cos{\theta}=4\cos^3{\theta}-3\cos{\theta}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
এবং \(\cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{\theta}(4\cos^2{\theta}-3)\)
\(\Rightarrow 2\sin{\theta}=4\cos^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=4(1-\sin^2{\theta})-3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2\sin{\theta}=4-4\sin^2{\theta}-3\)
\(\Rightarrow 2\sin{\theta}=1-4\sin^2{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+2\sin{\theta}-1=0\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{2^2-4\times4\times(-1)}}}{2\times4}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4+16}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{20}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{\sqrt{4\times5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-2\pm{2\sqrt{5}}}{8}\)
\(\Rightarrow \sin{\theta}=\frac{2(-1\pm{\sqrt{5}})}{8}\)
\(\Rightarrow \sin{\theta}=\frac{-1\pm{\sqrt{5}}}{4}\)
\(\Rightarrow \sin{18^{o}}=\frac{-1+\sqrt{5}}{4}\) ➜ \(\because 18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
তাই \(\sin{18^{o}}\) এর মান ধনাত্মক হবে।
\(\therefore \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\((b)\)
লেখা যায়,
\(\cos{36^{o}}=\sqrt{1-\sin^2{36^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(36^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}\sqrt{10-2\sqrt{5}}\right\}^2}\) ➜ \(\because \sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(=\sqrt{1-\frac{1}{16}(10-2\sqrt{5})}\)
\(=\sqrt{\frac{1}{16}\{16-(10-2\sqrt{5})\}}\)
\(=\frac{1}{4}\sqrt{16-10+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{5+1+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{(\sqrt{5}+1)^2}\) ➜ \(\because a^2+b^2+2ab=(a+b)^2\)
\(=\frac{1}{4}(\sqrt{5}+1)\)
\(\therefore \cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\(\cos{36^{o}}=\frac{1}{4}(\sqrt{5}+1)\)
\((c)\)
লেখা যায়,
\(\cos{18^{o}}=\sqrt{1-\sin^2{18^{o}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
এবং \(18^{o}\) কোণ ধনাত্মক ও সূক্ষ্ণ।
\(=\sqrt{1-\left\{\frac{1}{4}(\sqrt{5}-1)\right\}^2}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(=\sqrt{1-\frac{1}{16}(\sqrt{5}-1)^2}\)
\(=\sqrt{\frac{1}{16}\{16-(\sqrt{5}-1)^2\}}\)
\(=\frac{1}{4}\sqrt{16-(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{4}\sqrt{16-(6-2\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{16-6+2\sqrt{5}}\)
\(=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\therefore \cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\((d)\)
লেখা যায়,
\(\sin{36^{o}}=\sin{(2\times18^{o})}\)
\(=2\sin{18^{o}}\cos{18^{o}}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=2\times\frac{1}{4}(\sqrt{5}-1)\times\frac{1}{4}\sqrt{10+2\sqrt{5}}\) ➜ \(\because \sin{18^{o}}=\frac{1}{4}(\sqrt{5}-1)\)
\(\cos{18^{o}}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(=\frac{1}{8}(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}\)
\(=\frac{1}{8}(\sqrt{5}-1)\sqrt{2\sqrt{5}(\sqrt{5}+1)}\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(\sqrt{5}-1)^2}\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(5+1-2\sqrt{5})}\) ➜ \(\because (a-b)^2=a^2+b^2-2ab\)
\(=\frac{1}{8}\sqrt{2\sqrt{5}(\sqrt{5}+1)(6-2\sqrt{5})}\)
\(=\frac{1}{8}\sqrt{4\sqrt{5}(\sqrt{5}+1)(3-\sqrt{5})}\)
\(=\frac{2}{8}\sqrt{\sqrt{5}(3\sqrt{5}+3-5-\sqrt{5})}\)
\(=\frac{1}{4}\sqrt{\sqrt{5}(2\sqrt{5}-2)}\)
\(=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(\therefore \sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
\(\sin{36^{o}}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)
অধ্যায় \(7E\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
প্রমাণ করঃ
\(Q.1.(i)\) \(\frac{1+\sin{x}-\cos{x}}{1+\sin{x}+\cos{x}}=\tan{\frac{x}{2}}\)\(Q.1.(ii)\) \(\frac{1+\sin{\theta}}{1-\sin{\theta}}=\tan^2{\left(\frac{\pi}{4}+\frac{\theta}{2}\right)}\)
\(Q.1.(iii)\) \(\sec{x}+\tan{x}=\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}\)
\(Q.1.(iv)\) \(\cot{\theta}=\frac{1}{2}\left(\cot{\frac{\theta}{2}}-\tan{\frac{\theta}{2}}\right)\)
\(Q.1.(v)\) \(\frac{\cos{\frac{\theta}{2}}-\sqrt{1+\sin{\theta}}}{\sin{\frac{\theta}{2}}-\sqrt{1+\sin{\theta}}}=\tan{\frac{\theta}{2}}\)
\(Q.1.(vi)\) \(\frac{2\cos{A}-\cos{2A}-1}{2\cos{A}+\cos{2A}+1}=\tan^2{\frac{A}{2}}\)
প্রমাণ করঃ
\(Q.1.(vii)\) \(\frac{1-\tan{\frac{A}{2}}}{1+\tan{\frac{A}{2}}}=\frac{\cos{A}}{1+\sin{A}}\)\(Q.1.(viii)\) \(\frac{\sec{\alpha}-\tan{\alpha}}{\sec{\alpha}+\tan{\alpha}}=\cot^2{\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}\)
\(Q.1.(ix)\) \(\tan{\frac{\theta}{2}}+\cot{\frac{\theta}{2}}=2cosec \ {\theta}\)
\(Q.1.(x)\) \(\left(1+\tan{\frac{\alpha}{2}}-\sec{\frac{\alpha}{2}}\right)\left(1+\tan{\frac{\alpha}{2}}+\sec{\frac{\alpha}{2}}\right)=\sin{\alpha}\sec^2{\frac{\alpha}{2}}\)
\(Q.1.(xi)\) \(\frac{2\sin{\theta}-\sin{2\theta}}{2\sin{\theta}+\sin{2\theta}}=\tan^2{\frac{\theta}{2}}\)
অধ্যায় \(7E\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
প্রমাণ করঃ
\(Q.2.(i)\) \(\cos^4{\frac{\theta}{2}}+\sin^4{\frac{\theta}{2}}=\frac{1}{4}(3+\cos{2\theta})\)\(Q.2.(ii)\) \(\cos^2{\frac{\theta}{2}}+\cos^2{\left(\frac{\theta}{2}+60^{o}\right)}+\cos^2{\left(\frac{\theta}{2}-60^{o}\right)}=\frac{3}{2}\)
\(Q.2.(iii)\) \(\sin^2{\frac{\theta}{2}}+\sin^2{\left(\frac{\theta}{2}+60^{o}\right)}+\sin^2{\left(\frac{\theta}{2}-60^{o}\right)}=\frac{3}{2}\)
\(Q.2.(iv)\) \(\cos^4{\frac{\pi}{8}}+\cos^4{\frac{3\pi}{8}}+\cos^4{\frac{5\pi}{8}}+\cos^4{\frac{7\pi}{8}}=\frac{3}{2}\)
\(Q.2.(v)\) \(\sin^4{\frac{\pi}{8}}+\sin^4{\frac{3\pi}{8}}+\sin^4{\frac{5\pi}{8}}+\sin^4{\frac{7\pi}{8}}=\frac{3}{2}\)
\(Q.2.(vi)\) \(\sin^2{\frac{A}{2}}+\sin^2{\left(\frac{A}{2}+\frac{\pi}{3}\right)}+\sin^2{\left(\frac{A}{2}-\frac{\pi}{3}\right)}=\frac{3}{2}\)
প্রমাণ করঃ
\(Q.2.(vii)\) \(\sin^2{\left(\frac{\alpha}{2}-36^{o}\right)}+\sin^2{\left(\frac{\alpha}{2}+36^{o}\right)}=\frac{1}{4}\left\{4-(\sqrt{5}-1)\cos{\alpha}\right\}\)\(Q.2.(viii)\) \(\cos^2{\left(\frac{\alpha}{2}-18^{o}\right)}+\cos^2{\left(\frac{\alpha}{2}+18^{o}\right)}=\frac{1}{4}\left\{4+(\sqrt{5}+1)\cos{\alpha}\right\}\)
\(Q.2.(ix)\) \(\cos{2\theta}=8\cos^4{\frac{\theta}{2}}-8\cos^2{\frac{\theta}{2}}+1\)
\(Q.2.(x)\) \(\cos{2\theta}=8\sin^4{\frac{\theta}{2}}-8\sin^2{\frac{\theta}{2}}+1\)
\(Q.2.(xi)\) \(\cos^2{\frac{\theta}{2}}\left(1+\tan{\frac{\theta}{2}}\right)^2=1+\sin{\theta}\)
অধ্যায় \(7E\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
প্রমাণ করঃ
\(Q.3.(i)\) \(2\cos{\frac{\pi}{16}}=\sqrt{2+\sqrt{2+\sqrt{2}}}\) কুঃ ২০১৩,২০০৭।
\(Q.3.(ii)\) \(2\cos{\left(7\frac{1}{2}\right)^{o}}=\sqrt{2+\sqrt{2+\sqrt{3}}}\)
কুঃ, চঃ ২০১০; রাঃ২০০৩।
\(Q.3.(iii)\) \((\cos{\alpha}+\cos{\beta})^2+(\sin{\alpha}-\sin{\beta})^2=4\cos^2{\frac{\alpha+\beta}{2}}\)
যঃ ২০১২।
\(Q.3.(iv)\) \(\tan{6^{o}}\tan{42^{o}}\tan{66^{o}}\tan{78^{o}}=1\)
\(Q.3.(v)\) \(\tan{\left(7\frac{1}{2}\right)^{o}}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2\)
প্রমাণ করঃ
\(Q.3.(vi)\) \(\tan{\left(82\frac{1}{2}\right)^{o}}=\sqrt{6}+\sqrt{3}+\sqrt{2}+2\)\(Q.3.(vii)\) \((\cos{\alpha}-\cos{\beta})^2+(\sin{\alpha}-\sin{\beta})^2=4\sin^2{\frac{(\alpha-\beta)}{2}}\)
যঃ ২০১২।
\(Q.3.(viii)\) \(\sin{(292.5)^{o}}=-\frac{1}{2}\sqrt{2+\sqrt{2}}\)
\(Q.3.(ix)\) \(\cot{(142.5)^{o}}=\sqrt{2}+\sqrt{3}-2-\sqrt{6}\)
\(Q.3.(x)\) প্রমাণ কর যে, \(2\sin{\left(\frac{\pi}{64}\right)}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\)
প্রমাণ করঃ
\(Q.3.(vi)\) \(\tan{\left(82\frac{1}{2}\right)^{o}}=\sqrt{6}+\sqrt{3}+\sqrt{2}+2\)সমাধানঃ
\(L.S=\tan{\left(82\frac{1}{2}\right)^{o}}\)
\(=\tan{\left(\frac{165^{o}}{2}\right)}\)
\(=\tan{\left(90^{o}\times1-\frac{15^{o}}{2}\right)}\)
\(=\cot{\left(\frac{15^{o}}{2}\right)}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(=\frac{\cos{\left(\frac{15^{o}}{2}\right)}}{\sin{\left(\frac{15^{o}}{2}\right)}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=\frac{2\cos^2{\left(\frac{15^{o}}{2}\right)}}{2\sin{\left(\frac{15^{o}}{2}\right)}\cos{\left(\frac{15^{o}}{2}\right)}}\) ➜ লব ও হরকে \(2\cos{\left(\frac{15^{o}}{2}\right)}\) দ্বারা গুণ করে,
\(=\frac{1+\cos{15^{o}}}{\sin{15^{o}}}\) ➜ \(\because 2\cos^2{\left(\frac{A}{2}\right)}=1+\cos{A}\)
এবং \(2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}=\sin{A}\)
\(=\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}\) ➜ \(\because \cos{15^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
এবং \(\sin{15^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\) ➜ লব ও হরকে \(2\sqrt{2}\) দ্বারা গুণ করে,
\(=\frac{(\sqrt{3}+1)(2\sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\) ➜ আবার, লব ও হরকে \((\sqrt{3}+1)\) দ্বারা গুণ করে,
\(=\frac{2\sqrt{6}+3+\sqrt{3}+2\sqrt{2}+\sqrt{3}+1}{3-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{2\sqrt{6}+2\sqrt{3}+2\sqrt{2}+4}{2}\)
\(=\frac{2(\sqrt{6}+\sqrt{3}+\sqrt{2}+2)}{2}\)
\(=\sqrt{6}+\sqrt{3}+\sqrt{2}+2\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(=\tan{\left(\frac{165^{o}}{2}\right)}\)
\(=\tan{\left(90^{o}\times1-\frac{15^{o}}{2}\right)}\)
\(=\cot{\left(\frac{15^{o}}{2}\right)}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(=\frac{\cos{\left(\frac{15^{o}}{2}\right)}}{\sin{\left(\frac{15^{o}}{2}\right)}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=\frac{2\cos^2{\left(\frac{15^{o}}{2}\right)}}{2\sin{\left(\frac{15^{o}}{2}\right)}\cos{\left(\frac{15^{o}}{2}\right)}}\) ➜ লব ও হরকে \(2\cos{\left(\frac{15^{o}}{2}\right)}\) দ্বারা গুণ করে,
\(=\frac{1+\cos{15^{o}}}{\sin{15^{o}}}\) ➜ \(\because 2\cos^2{\left(\frac{A}{2}\right)}=1+\cos{A}\)
এবং \(2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}=\sin{A}\)
\(=\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}\) ➜ \(\because \cos{15^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
এবং \(\sin{15^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\) ➜ লব ও হরকে \(2\sqrt{2}\) দ্বারা গুণ করে,
\(=\frac{(\sqrt{3}+1)(2\sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\) ➜ আবার, লব ও হরকে \((\sqrt{3}+1)\) দ্বারা গুণ করে,
\(=\frac{2\sqrt{6}+3+\sqrt{3}+2\sqrt{2}+\sqrt{3}+1}{3-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\frac{2\sqrt{6}+2\sqrt{3}+2\sqrt{2}+4}{2}\)
\(=\frac{2(\sqrt{6}+\sqrt{3}+\sqrt{2}+2)}{2}\)
\(=\sqrt{6}+\sqrt{3}+\sqrt{2}+2\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
প্রমাণ করঃ
\(Q.3.(viii)\) \(\sin{(292.5)^{o}}=-\frac{1}{2}\sqrt{2+\sqrt{2}}\)সমাধানঃ
\(L.S=\sin{(292.5)^{o}}\)
\(=\sin{\{90^{o}\times3+(22.5)^{o}\}}\) ➜ \(\because (292.5)^{o}=90^{o}\times3+(22.5)^{o}\)
\(=-\cos{(22.5)^{o}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি চতুর্থ চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(3\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(=-\frac{1}{2}\sqrt{4\cos^2{(22.5)^{o}}}\)
\(=-\frac{1}{2}\sqrt{2\times2\cos^2{(22.5)^{o}}}\)
\(=-\frac{1}{2}\sqrt{2\{1+\cos{2\times(22.5)^{o}}\}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=-\frac{1}{2}\sqrt{2+2\cos{45^{o}}}\)
\(=-\frac{1}{2}\sqrt{2+2\times\frac{1}{\sqrt{2}}}\) ➜ \(\because \cos{45^{o}}=\frac{1}{\sqrt{2}}\)
\(=-\frac{1}{2}\sqrt{2+\frac{2}{\sqrt{2}}}\)
\(=-\frac{1}{2}\sqrt{2+\frac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}}\)
\(=-\frac{1}{2}\sqrt{2+\sqrt{2}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(=\sin{\{90^{o}\times3+(22.5)^{o}\}}\) ➜ \(\because (292.5)^{o}=90^{o}\times3+(22.5)^{o}\)
\(=-\cos{(22.5)^{o}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি চতুর্থ চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(3\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(=-\frac{1}{2}\sqrt{4\cos^2{(22.5)^{o}}}\)
\(=-\frac{1}{2}\sqrt{2\times2\cos^2{(22.5)^{o}}}\)
\(=-\frac{1}{2}\sqrt{2\{1+\cos{2\times(22.5)^{o}}\}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=-\frac{1}{2}\sqrt{2+2\cos{45^{o}}}\)
\(=-\frac{1}{2}\sqrt{2+2\times\frac{1}{\sqrt{2}}}\) ➜ \(\because \cos{45^{o}}=\frac{1}{\sqrt{2}}\)
\(=-\frac{1}{2}\sqrt{2+\frac{2}{\sqrt{2}}}\)
\(=-\frac{1}{2}\sqrt{2+\frac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}}\)
\(=-\frac{1}{2}\sqrt{2+\sqrt{2}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
প্রমাণ করঃ
\(Q.3.(ix)\) \(\cot{(142.5)^{o}}=\sqrt{2}+\sqrt{3}-2-\sqrt{6}\)সমাধানঃ
\(L.S=\cot{(142.5)^{o}}\)
\(=\cot{\{90^{o}\times2-(37.5)^{o}\}}\) ➜ \(\because (142.5)^{o}=90^{o}\times2-(37.5)^{o}\)
\(=-\cot{(37.5)^{o}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
সুতরাং অনুপাতের পরিবর্তন হয়নি।
\(=-\frac{\cos{(37.5)^{o}}}{\sin{(37.5)^{o}}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=-\frac{2\cos^2{(37.5)^{o}}}{2\sin{(37.5)^{o}}\cos{(37.5)^{o}}}\) ➜ লব ও হরকে \(2\cos{(37.5)^{o}}\) দ্বারা গুন করে,
\(=-\frac{1+\cos{2\times(37.5)^{o}}}{\sin{2\times(37.5)^{o}}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(=-\frac{1+\cos{75^{o}}}{\sin{75^{o}}}\)
\(=-\frac{1+\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\) ➜ \(\because \cos{75^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
এবং \(\sin{75^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
\(=-\frac{2\sqrt{2}+\sqrt{3}-1}{\sqrt{3}+1}\) ➜ লব ও হরকে \(2\sqrt{2}\) দ্বারা গুন করে,
\(=-\frac{(\sqrt{3}-1)(2\sqrt{2}+\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\) ➜ আবার, লব ও হরকে \((\sqrt{3}-1)\) দ্বারা গুন করে,
\(=-\frac{2\sqrt{6}+3-\sqrt{3}-2\sqrt{2}-\sqrt{3}+1}{3-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=-\frac{2\sqrt{6}-2\sqrt{3}-2\sqrt{2}+4}{2}\)
\(=-\frac{2(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)}{2}\)
\(=\sqrt{2}+\sqrt{3}-2-\sqrt{6}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(=\cot{\{90^{o}\times2-(37.5)^{o}\}}\) ➜ \(\because (142.5)^{o}=90^{o}\times2-(37.5)^{o}\)
\(=-\cot{(37.5)^{o}}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
সুতরাং অনুপাতের পরিবর্তন হয়নি।
\(=-\frac{\cos{(37.5)^{o}}}{\sin{(37.5)^{o}}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=-\frac{2\cos^2{(37.5)^{o}}}{2\sin{(37.5)^{o}}\cos{(37.5)^{o}}}\) ➜ লব ও হরকে \(2\cos{(37.5)^{o}}\) দ্বারা গুন করে,
\(=-\frac{1+\cos{2\times(37.5)^{o}}}{\sin{2\times(37.5)^{o}}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(=-\frac{1+\cos{75^{o}}}{\sin{75^{o}}}\)
\(=-\frac{1+\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\) ➜ \(\because \cos{75^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
এবং \(\sin{75^{o}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
\(=-\frac{2\sqrt{2}+\sqrt{3}-1}{\sqrt{3}+1}\) ➜ লব ও হরকে \(2\sqrt{2}\) দ্বারা গুন করে,
\(=-\frac{(\sqrt{3}-1)(2\sqrt{2}+\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\) ➜ আবার, লব ও হরকে \((\sqrt{3}-1)\) দ্বারা গুন করে,
\(=-\frac{2\sqrt{6}+3-\sqrt{3}-2\sqrt{2}-\sqrt{3}+1}{3-1}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=-\frac{2\sqrt{6}-2\sqrt{3}-2\sqrt{2}+4}{2}\)
\(=-\frac{2(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)}{2}\)
\(=\sqrt{2}+\sqrt{3}-2-\sqrt{6}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
অধ্যায় \(7E\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
\(Q.4.(i)\) যদি \(\sin{\theta}+\sin{\phi}=a\) এবং \(\cos{\theta}+\cos{\phi}=b\) হয়, তবে দেখাও যে,
\((a) \ \cos{(\theta+\phi)}=\frac{b^2-a^2}{b^2+a^2}\)
\((c) \ \tan{\frac{1}{2}(\theta-\phi)}=\pm\frac{1}{2}\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}\)
\(Q.4.(ii)\) যদি \(a\sin{\alpha}+b\sin{\beta}=c\) এবং \(a\cos{\alpha}+b\cos{\beta}=c\) হয়, তবে প্রমাণ কর যে,
\((a) \ \sin{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{\frac{(a+b)^2-2c^2}{ab}}\)
\((b) \ \cos{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{\frac{2c^2-(a-b)^2}{ab}}\)
\(Q.4.(iii)\) \(\sqrt{1+n}\tan{\frac{\alpha}{2}}=\sqrt{1-n}\tan{\frac{\beta}{2}}\) তবে দেখাও যে, \(\cos{\beta}=\frac{\cos{\alpha}-n}{1-n\cos{\alpha}}\)
\(Q.4.(iv)\) যদি \(A+B\ne{0}\) এবং \(\sin{A}+\sin{B}=2\sin{(A+B)}\) হয়, তবে প্রমাণ কর যে, \(\tan{\frac{A}{2}}\tan{\frac{B}{2}}=\frac{1}{3}\)
\(Q.4.(v)\) \(\triangle{ABC}\) এর \(A=75^{o}\) ও \(B-C=15^{o}\) হলে দেখাও যে, \(\cos{\frac{C}{6}}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}\)
\(Q.4.(vi)\) \(\angle{E}+\angle{F}=65^{0}, \ \angle{F}-\angle{E}=25^{0}\) হলে দেখাও যে, \(2\sin{\left(\pi+\frac{F}{4}\right)}=-\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\((a) \ \cos{(\theta+\phi)}=\frac{b^2-a^2}{b^2+a^2}\)
রাঃ২০০৮,২০০৩; কুঃ২০১৪; সিঃ২০১৬,২০১১; বুয়েটঃ২০১১-২০১২।
\((b) \ \cos{\frac{1}{2}(\theta-\phi)}=\pm\frac{1}{2}\sqrt{a^2+b^2}\)\((c) \ \tan{\frac{1}{2}(\theta-\phi)}=\pm\frac{1}{2}\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}\)
কুঃ২০১১।
\(Q.4.(ii)\) যদি \(a\sin{\alpha}+b\sin{\beta}=c\) এবং \(a\cos{\alpha}+b\cos{\beta}=c\) হয়, তবে প্রমাণ কর যে,
\((a) \ \sin{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{\frac{(a+b)^2-2c^2}{ab}}\)
\((b) \ \cos{\frac{1}{2}(\alpha-\beta)}=\pm\frac{1}{2}\sqrt{\frac{2c^2-(a-b)^2}{ab}}\)
\(Q.4.(iii)\) \(\sqrt{1+n}\tan{\frac{\alpha}{2}}=\sqrt{1-n}\tan{\frac{\beta}{2}}\) তবে দেখাও যে, \(\cos{\beta}=\frac{\cos{\alpha}-n}{1-n\cos{\alpha}}\)
যঃ২০১৭।
\(Q.4.(iv)\) যদি \(A+B\ne{0}\) এবং \(\sin{A}+\sin{B}=2\sin{(A+B)}\) হয়, তবে প্রমাণ কর যে, \(\tan{\frac{A}{2}}\tan{\frac{B}{2}}=\frac{1}{3}\)
যঃ২০১৭।
\(Q.4.(v)\) \(\triangle{ABC}\) এর \(A=75^{o}\) ও \(B-C=15^{o}\) হলে দেখাও যে, \(\cos{\frac{C}{6}}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}\)
যঃ২০১৯; কুঃ,চঃ ২০১০।
\(Q.4.(vi)\) \(\angle{E}+\angle{F}=65^{0}, \ \angle{F}-\angle{E}=25^{0}\) হলে দেখাও যে, \(2\sin{\left(\pi+\frac{F}{4}\right)}=-\sqrt{2-\sqrt{2+\sqrt{2}}}\)
যঃ২০১৭।
\(Q.4.(vii)\) \(x=\sin{\frac{\pi}{18}}\) হলে দেখাও যে, \(8x^4+4x^3-6x^2-2x+\frac{1}{2}=0\)
\(Q.4.(viii)\) \(\cos{\theta}=\frac{a\cos{\phi}-b}{a-b\cos{\phi}}\) হলে দেখাও যে, \(\frac{\tan{\frac{1}{2}\theta}}{\sqrt{a+b}}=\frac{\tan{\frac{1}{2}\phi}}{\sqrt{a-b}}\)
\(Q.4.(ix)\) \(\cos{36^{o}}\) এর মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}(\sqrt{5}+1)\)
\(Q.4.(x)\) \(\sec{(\theta+\alpha)}+\sec{(\theta-\alpha)}=2\sec{\theta}\) হলে দেখাও যে, \(\cos{\theta}=\pm{\sqrt{2}\cos{\frac{\alpha}{2}}}\)
\(Q.4.(xi)\) \(\sin{A}=\frac{1}{\sqrt{2}}\) এবং \(\sin{B}=\frac{1}{\sqrt{3}}\) হলে দেখাও যে, \(\tan{\frac{A+B}{2}}\cot{\frac{A-B}{2}}=5+2\sqrt{6}\)
\(Q.4.(xii)\) \(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণটি \(\theta\) এর দুইটি ভিন্ন মান \(\alpha\) ও \(\beta\) দ্বারা সিদ্ধ হলে দেখাও যে, \(\sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
\(Q.4.(xiii)\) দেখাও যে, \(\sin{x}=2^n\cos{\frac{x}{2}}\cos{\frac{x}{2^2}}\cos{\frac{x}{2^3}} ...... \cos{\frac{x}{2^n}}\sin{\frac{x}{2^2}}\)
\(Q.4.(xiv)\) যদি \(\sin{\theta}=\frac{a-b}{a+b}\) হয় তবে প্রমাণ কর যে, \(\tan{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\pm{\sqrt{\frac{b}{a}}}\)
\(Q.4.(xv)\) যদি \(\cos{\theta}=\frac{\cos{\alpha}-\cos{\beta}}{1-\cos{\alpha}\cos{\beta}}\) হয় তবে প্রমাণ কর যে, \(\tan{\frac{\theta}{2}}\) অনুপাতের একটি মান হবে \(\tan{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\)
\(Q.4.(xvi)\) \(x\cos{\alpha}+y\sin{\alpha}=k=x\cos{\beta}+y\sin{\beta}\) হলে দেখাও যে, \(\frac{x}{\cos{\frac{1}{2}(\alpha+\beta)}}=\frac{y}{\sin{\frac{1}{2}(\alpha+\beta)}}=\frac{k}{\cos{\frac{1}{2}(\alpha-\beta)}}\)
\(Q.4.(viii)\) \(\cos{\theta}=\frac{a\cos{\phi}-b}{a-b\cos{\phi}}\) হলে দেখাও যে, \(\frac{\tan{\frac{1}{2}\theta}}{\sqrt{a+b}}=\frac{\tan{\frac{1}{2}\phi}}{\sqrt{a-b}}\)
\(Q.4.(ix)\) \(\cos{36^{o}}\) এর মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}(\sqrt{5}+1)\)
\(Q.4.(x)\) \(\sec{(\theta+\alpha)}+\sec{(\theta-\alpha)}=2\sec{\theta}\) হলে দেখাও যে, \(\cos{\theta}=\pm{\sqrt{2}\cos{\frac{\alpha}{2}}}\)
\(Q.4.(xi)\) \(\sin{A}=\frac{1}{\sqrt{2}}\) এবং \(\sin{B}=\frac{1}{\sqrt{3}}\) হলে দেখাও যে, \(\tan{\frac{A+B}{2}}\cot{\frac{A-B}{2}}=5+2\sqrt{6}\)
\(Q.4.(xii)\) \(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণটি \(\theta\) এর দুইটি ভিন্ন মান \(\alpha\) ও \(\beta\) দ্বারা সিদ্ধ হলে দেখাও যে, \(\sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
\(Q.4.(xiii)\) দেখাও যে, \(\sin{x}=2^n\cos{\frac{x}{2}}\cos{\frac{x}{2^2}}\cos{\frac{x}{2^3}} ...... \cos{\frac{x}{2^n}}\sin{\frac{x}{2^2}}\)
\(Q.4.(xiv)\) যদি \(\sin{\theta}=\frac{a-b}{a+b}\) হয় তবে প্রমাণ কর যে, \(\tan{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\pm{\sqrt{\frac{b}{a}}}\)
\(Q.4.(xv)\) যদি \(\cos{\theta}=\frac{\cos{\alpha}-\cos{\beta}}{1-\cos{\alpha}\cos{\beta}}\) হয় তবে প্রমাণ কর যে, \(\tan{\frac{\theta}{2}}\) অনুপাতের একটি মান হবে \(\tan{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\)
\(Q.4.(xvi)\) \(x\cos{\alpha}+y\sin{\alpha}=k=x\cos{\beta}+y\sin{\beta}\) হলে দেখাও যে, \(\frac{x}{\cos{\frac{1}{2}(\alpha+\beta)}}=\frac{y}{\sin{\frac{1}{2}(\alpha+\beta)}}=\frac{k}{\cos{\frac{1}{2}(\alpha-\beta)}}\)
\(Q.4.(vi)\) \(\angle{E}+\angle{F}=65^{0}, \ \angle{F}-\angle{E}=25^{0}\) হলে দেখাও যে, \(2\sin{\left(\pi+\frac{F}{4}\right)}=-\sqrt{2-\sqrt{2+\sqrt{2}}}\)
যঃ২০১৭।
সমাধানঃ
দেওয়া আছে,
\(\angle{E}+\angle{F}=65^{o}, \ \angle{F}-\angle{E}=25^{o}\)
ধরি,
\(\angle{E}+\angle{F}=65^{o}........(1)\)
\(\angle{F}-\angle{E}=25^{o} ........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(\angle{E}+\angle{F}+\angle{F}-\angle{E}=65^{o}+25^{o}\)
\(\Rightarrow 2\angle{F}=90^{o}\)
\(\Rightarrow \angle{F}=\frac{90^{o}}{2}\)
\(\therefore \angle{F}=45^{o}\)
\(L.S=2\sin{\left(\pi+\frac{F}{4}\right)}\)
\(=2\sin{\left(\frac{\pi}{2}\times2+\frac{45^{o}}{4}\right)}\)
\(=-2\sin{\left(\frac{45^{o}}{4}\right)}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।
\(=-2\sin{11^{o}15^{\prime}}\)
\(=-\sqrt{4\sin^2{\left(\frac{45^{o}}{4}\right)}}\)
\(=-\sqrt{2\times2\sin^2{\left(\frac{45^{o}}{4}\right)}}\)
\(=-\sqrt{2\left\{1-\cos{2\left(\frac{45^{o}}{4}\right)}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=-\sqrt{2\left\{1-\cos{\left(\frac{45^{o}}{2}\right)}\right\}}\)
\(=-\sqrt{2-2\cos{\left(\frac{45^{o}}{2}\right)}}\)
\(=-\sqrt{2-\sqrt{4\cos^2{\left(\frac{45^{o}}{2}\right)}}}\)
\(=-\sqrt{2-\sqrt{2\times2\cos^2{\left(\frac{45^{o}}{2}\right)}}}\)
\(=-\sqrt{2-\sqrt{2\left\{1+\cos{2\left(\frac{45^{o}}{2}\right)}\right\}}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=-\sqrt{2-\sqrt{2\left\{1+\cos{\left(45^{o}\right)}\right\}}}\)
\(=-\sqrt{2-\sqrt{2+2\cos{\left(45^{o}\right)}}}\)
\(=-\sqrt{2-\sqrt{2+2\times\frac{1}{\sqrt{2}}}}\) ➜ \(\because \cos{\left(45^{o}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\sqrt{2-\sqrt{2+\frac{2}{\sqrt{2}}}}\)
\(=-\sqrt{2-\sqrt{2+\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}}}\)
\(=-\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(\angle{E}+\angle{F}=65^{o}, \ \angle{F}-\angle{E}=25^{o}\)
ধরি,
\(\angle{E}+\angle{F}=65^{o}........(1)\)
\(\angle{F}-\angle{E}=25^{o} ........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(\angle{E}+\angle{F}+\angle{F}-\angle{E}=65^{o}+25^{o}\)
\(\Rightarrow 2\angle{F}=90^{o}\)
\(\Rightarrow \angle{F}=\frac{90^{o}}{2}\)
\(\therefore \angle{F}=45^{o}\)
\(L.S=2\sin{\left(\pi+\frac{F}{4}\right)}\)
\(=2\sin{\left(\frac{\pi}{2}\times2+\frac{45^{o}}{4}\right)}\)
\(=-2\sin{\left(\frac{45^{o}}{4}\right)}\) ➜
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।
\(=-2\sin{11^{o}15^{\prime}}\)
\(=-\sqrt{4\sin^2{\left(\frac{45^{o}}{4}\right)}}\)
\(=-\sqrt{2\times2\sin^2{\left(\frac{45^{o}}{4}\right)}}\)
\(=-\sqrt{2\left\{1-\cos{2\left(\frac{45^{o}}{4}\right)}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=-\sqrt{2\left\{1-\cos{\left(\frac{45^{o}}{2}\right)}\right\}}\)
\(=-\sqrt{2-2\cos{\left(\frac{45^{o}}{2}\right)}}\)
\(=-\sqrt{2-\sqrt{4\cos^2{\left(\frac{45^{o}}{2}\right)}}}\)
\(=-\sqrt{2-\sqrt{2\times2\cos^2{\left(\frac{45^{o}}{2}\right)}}}\)
\(=-\sqrt{2-\sqrt{2\left\{1+\cos{2\left(\frac{45^{o}}{2}\right)}\right\}}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=-\sqrt{2-\sqrt{2\left\{1+\cos{\left(45^{o}\right)}\right\}}}\)
\(=-\sqrt{2-\sqrt{2+2\cos{\left(45^{o}\right)}}}\)
\(=-\sqrt{2-\sqrt{2+2\times\frac{1}{\sqrt{2}}}}\) ➜ \(\because \cos{\left(45^{o}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\sqrt{2-\sqrt{2+\frac{2}{\sqrt{2}}}}\)
\(=-\sqrt{2-\sqrt{2+\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}}}\)
\(=-\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
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