Question

The value of $x\in (0,{90}^{\circ })$ satisfying $cosx=sin{61}^{\circ }+sin{47}^{\circ }-sin{25}^{\circ }-sin{11}^{\circ }$, where $\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}$

$\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

• A. $7^{\circ }$
• B. ${11}^{\circ }$
• C. ${13}^{\circ }$
• D. ${17}^{\circ }$

Answer 1

The correct option is A $7^{\circ }$

Consider the given equation.

$\cos x=\sin {61}^{\circ }+\sin {47}^{\circ }-\sin {25}^{\circ }-\sin {11}^{\circ }$

$\cos x=\sin {47}^{\circ }-\sin {25}^{\circ }+\sin {61}^{\circ }-\sin {11}^{\circ }$

We know that

$\sin C-\mathrm{sinD}=2cos\left(\frac{C+D}{2}\right)\cdot \sin \left(\frac{C-D}{2}\right)$

Therefore,

$\cos x=2cos\left(\frac{{47}^{\circ }+{25}^{\circ }}{2}\right)\cdot \sin \left(\frac{{47}^{\circ }-{25}^{\circ }}{2}\right)+2cos\left(\frac{{61}^{\circ }+{11}^{\circ }}{2}\right)\cdot \sin \left(\frac{{61}^{\circ }-{11}^{\circ }}{2}\right)$

$\cos x=2cos\left(\frac{{72}^{\circ }}{2}\right)\cdot \sin \left(\frac{{22}^{\circ }}{2}\right)+2cos\left(\frac{{72}^{\circ }}{2}\right)\cdot \sin \left(\frac{{50}^{\circ }}{2}\right)$

$\cos x=2cos\left({36}^{\circ }\right)\cdot \sin \left({11}^{\circ }\right)+2cos\left({36}^{\circ }\right)\cdot \sin \left({25}^{\circ }\right)$

$\cos x=2cos\left({36}^{\circ }\right)[\sin \left({25}^{\circ }\right)+\sin \left({11}^{\circ }\right)]$

$\cos x=2cos\left({36}^{\circ }\right)[2\sin \left(\frac{{25}^{\circ }+{11}^{\circ }}{2}\right)\cdot \cos \left(\frac{{25}^{\circ }-{11}^{\circ }}{2}\right)]$

$\cos x=4cos\left({36}^{\circ }\right)\sin \left({18}^{\circ }\right)\cdot \cos \left(7^{\circ }\right)$

Since,

$\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}$

$\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

Therefore,

$\cos x=4\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right)\cdot \cos \left(7^{\circ }\right)$

$\cos x=\left(\frac{5-1}{4}\right)\cdot \cos \left(7^{\circ }\right)$

$\cos x=\left(\frac{4}{4}\right)\cdot \cos \left(7^{\circ }\right)$

$\cos x=\cos \left(7^{\circ }\right)$

$x=7^{\circ }$

Hence, this is the answer.