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If 

\cos A = \frac{\sqrt{10-2\sqrt{5}}}{4} and \cos B = \frac{\sqrt{5}+1}{4} then the value of \cos(\frac{ A-B}{2})

Option: 1

\frac{{10-\sqrt{5}}+1}{4\sqrt{2}}\\


Option: 2

\frac{\sqrt{10-2\sqrt{5}}-\sqrt{5}11}{4\sqrt{2}}\\


Option: 3

\frac{\sqrt{10-2\sqrt{5}}+\sqrt{5}+1}{4\sqrt{2}}\\


Option: 4

\frac{\sqrt{10-2\sqrt{5}}+\sqrt{5}+1}{4}\\


Answers (1)

best_answer

Value Trigonometric Ratio of some Particular Angle (Applications) (Part 2) -

Value Trigonometric Ratio of some Particular Angle(Applications) (Part 2)

 

4. cos 360

\\\mathrm{We \;know\;that,\;cos\;2\theta=1-\2\sin^2\theta}\\\mathrm{Put\;\theta=18^{\circ}}\\\mathrm{\cos\;2\times18^{\circ}=1-2\sin^2(18^{\circ})}\\\mathrm{\cos\;36^{\circ}=1-2\sin^2(18^{\circ})}\\\mathrm{\cos\;36^{\circ}=1-2\times\left ( \frac{\sqrt5-1}{4} \right )^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;=1-2\times\frac{5-2\sqrt5+1}{16}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;=1-\frac{3-\sqrt5}{4}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;=\frac{4-3+\sqrt5}{4}=\frac{\sqrt5+1}{4}}\\\mathrm{Hence,\cos 36^{\circ}=\frac{\sqrt5+1}{4}}

5. sin 360

\\\mathrm{\sin^2\theta+\cos^2\theta=1}\\\mathrm{\sin^2\theta=1-\cos^2\theta}\\\mathrm{\sin^2(36^{\circ})=1-\cos^2(36^{\circ})}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1-\left ( \frac{\sqrt5+1}{4} \right )^2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1- \frac{6+2\sqrt5}{16}=\frac{16-6-2\sqrt5}{16}=\frac{10-2\sqrt5}{16}}\\\\\mathrm{\therefore \sin36^{\circ}=\frac{\sqrt{10-2\sqrt5}}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;[\because \sin36^{\circ}>0]}

6.  sin 540 and  cos 540
\\\mathrm{\sin 54^{\circ}=\sin\left ( 90^{\circ}-36^{\circ} \right )=\cos36^{\circ}}\\\\\mathrm{\therefore \sin54^{\circ}=\frac{\sqrt5+1}{4}}\\\\\\\mathrm{\cos54^{\circ}=\cos\left ( 90^{\circ}-36^{\circ} \right )=\sin36^{\circ}}\\\\\mathrm{\therefore \cos36^{\circ}=\frac{1}{4}\sqrt{10-2\sqrt5}}

7. cos 22.50

\\\mathrm{Let\;\;\theta=22.5^{\circ},\;then\;2\theta=45^{\circ}}\\\mathrm{Use\;the\;Identity,\;\;\cos2\theta=2\cos^2\theta-1}\\\mathrm{\cos^2\left ( 22.5^{\circ} \right )=\frac{1+\cos\left ( 45^{\circ} \right )}{2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1+\frac{1}{\sqrt2}}{2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\sqrt2+1}{2\sqrt2}=\frac{2+\sqrt2}{4}}\\\\\mathrm{\therefore \cos22.5^{\circ}=\frac{1}{2}\sqrt{2+\sqrt2}}

8. sin 22.50

\\\mathrm{Let\;\;\theta=22.5^{\circ},\;then\;2\theta=45^{\circ}}\\\mathrm{Use\;the\;Identity,\;\;\cos2\theta=1-2\sin^2\theta}\\\mathrm{\sin^2\left ( 22.5^{\circ} \right )=\frac{1-\cos\left ( 45^{\circ} \right )}{2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1-\frac{1}{\sqrt2}}{2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\sqrt2-1}{2\sqrt2}=\frac{2-\sqrt2}{4}}\\\\\mathrm{\therefore \sin22.5^{\circ}=\frac{1}{2}\sqrt{2-\sqrt2}}

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\cos A = \frac{\sqrt{10-2\sqrt{5}}}{4}\Rightarrow A=36^{\circ} and \cos B = \frac{\sqrt{5}+1}{4}\Rightarrow B=54^{\circ}

\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\Rightarrow take\ \alpha=36\ and\ \beta=54\\ \frac{\sqrt{10-2\sqrt{5}}}{4} + \frac{\sqrt{5}+1}{4}=2 \cos{45^{\circ}} \cdot \cos(9^{\circ})\\ \cos(\frac{A-B}{2})=\frac{\sqrt{10-2\sqrt{5}}+\sqrt{5}+1}{4\sqrt{2}}\\

Posted by

Divya Prakash Singh

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