Question

$sin{47}^{\circ }+sin{61}^{\circ }-sin{11}^{\circ }-sin{25}^{\circ }$ is equal to

• A. $sin{36}^{\circ }$
• B. $cos{36}^{\circ }$
• C. $sin{7}^{\circ }$
• D. $cos{7}^{\circ }$

Answer 1

The correct option is D

$cos{7}^{\circ }$

$sin{47}^{\circ }+sin{61}^{\circ }-sin{11}^{\circ }-sin{25}^{\circ }sin{47}^{\circ }-sin{25}^{\circ }+sin{61}^{\circ }-sin{11}^{\circ }=2sin\left(\frac{{47}^{\circ }-{25}^{\circ }}{2}\right)cos\left(\frac{{47}^{\circ }+{25}^{\circ }}{2}\right)+2sin\left(\frac{{61}^{\circ }-{11}^{\circ }}{2}\right)cos\left(\frac{{61}^{\circ }+{11}^{\circ }}{2}\right)=2sin{11}^{\circ }cos{36}^{\circ }+2sin{25}^{\circ }cos{36}^{\circ }=2cos{36}^{\circ }(sin{11}^{\circ }+sin{25}^{\circ })=2cos{36}^{\circ }\left\{2sin\left(\frac{{11}^{\circ }+{25}^{\circ }}{2}\right)cos\left(\frac{{11}^{\circ }-{25}^{\circ }}{2}\right)\right\}=4cos{36}^{\circ }sin{18}^{\circ }cos{7}^{\circ }=4\times \left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)cos{7}^{\circ }[cos{36}^{\circ }=\frac{\sqrt{5}+1}{4}andsin{18}^{\circ }\frac{\sqrt{5}-1}{4}]=\frac{5-1}{4}cos{7}^{\circ }=cos{7}^{\circ }$