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 $cos{7}^{\circ }$

$\sin {47}^{\circ }+\sin {61}^{\circ }-(\sin {11}^{\circ }+\sin {25}^{\circ })$

Using the transformation angle formula $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

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

$=\left(2\sin \left(\frac{108}{2}\right)\cos \left(\frac{-14}{2}\right)\right)-\left(2\sin \left(\frac{36}{2}\right)\cos \left(\frac{-14}{2}\right)\right)$

$=\left(2\sin {54}^{\circ }\cos {-7}^{\circ }\right)-\left(2\sin {18}^{\circ }\cos {-7}^{\circ }\right)$

$=2\cos 7^{\circ }(\sin {54}^{\circ }-\sin {18}^{\circ })$ using $\cos (-\theta )=\cos \theta$

$=2\cos 7^{\circ }\left(2\sin \left(\frac{54-18}{2}\right)\cos \left(\frac{54+18}{2}\right)\right)$ using the transformation angle formula $\sin C-\sin D=2\sin \left(\frac{C-D}{2}\right)\cos \left(\frac{C+D}{2}\right)$

$=4\cos 7^{\circ }\sin {18}^{\circ }\cos {36}^{\circ }$

We know that $\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}$ and $\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

$=4\cos 7^{\circ }\times \frac{\sqrt{5}-1}{4}\times \frac{\sqrt{5}+1}{4}$

$=4\cos 7^{\circ }\times \frac{5-1}{4\times 4}$

$=4\cos 7^{\circ }\times \frac{4}{4\times 4}=\cos 7^{\circ }$