Question

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

A
sin36
B
cos36
C
sin7
D
cos7

Solution

## The correct option is D $$cos\:7^{\circ }$$$$\sin{{47}^{\circ}}+\sin{{61}^{\circ}}-\left(\sin{{11}^{\circ}}+\sin{{25}^{\circ}}\right)$$ Using the transformation angle formula $$\sin{C}+\sin{D}=2\sin{\left(\dfrac{C+D}{2}\right)}\cos{\left(\dfrac{C-D}{2}\right)}$$ $$=\left(2\sin{\left(\dfrac{47+61}{2}\right)}\cos{\left(\dfrac{47-61}{2}\right)}\right)-\left(2\sin{\left(\dfrac{11+25}{2}\right)}\cos{\left(\dfrac{11-25}{2}\right)}\right)$$$$=\left(2\sin{\left(\dfrac{108}{2}\right)}\cos{\left(\dfrac{-14}{2}\right)}\right)-\left(2\sin{\left(\dfrac{36}{2}\right)}\cos{\left(\dfrac{-14}{2}\right)}\right)$$ $$=\left(2\sin{{54}^{\circ}}\cos{{-7}^{\circ}}\right)-\left(2\sin{{18}^{\circ}}\cos{{-7}^{\circ}}\right)$$$$=2\cos{{7}^{\circ}}\left(\sin{{54}^{\circ}}-\sin{{18}^{\circ}}\right)$$ using $$\cos{\left(-\theta\right)}=\cos{\theta}$$$$=2\cos{{7}^{\circ}}\left(2\sin{\left(\dfrac{54-18}{2}\right)}\cos{\left(\dfrac{54+18}{2}\right)}\right)$$ using the transformation angle formula $$\sin{C}-\sin{D}=2\sin{\left(\dfrac{C-D}{2}\right)}\cos{\left(\dfrac{C+D}{2}\right)}$$ $$=4\cos{{7}^{\circ}}\sin{{18}^{\circ}}\cos{{36}^{\circ}}$$We know that $$\sin{{18}^{\circ}}=\dfrac{\sqrt{5}-1}{4}$$ and $$\cos{{36}^{\circ}}=\dfrac{\sqrt{5}+1}{4}$$$$=4\cos{{7}^{\circ}}\times \dfrac{\sqrt{5}-1}{4}\times \dfrac{\sqrt{5}+1}{4}$$$$=4\cos{{7}^{\circ}}\times \dfrac{5-1}{4\times 4}$$$$=4\cos{{7}^{\circ}}\times \dfrac{4}{4\times 4}=\cos{{7}^{\circ}}$$Mathematics

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