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Question

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


A
sin36
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B
cos36
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C
sin7
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D
cos7
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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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