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Question

Mathematics Question on Inverse Trigonometric Functions

sin⁡  47o+sin⁡61o−sin⁡11o−sin⁡25o\sin \,\,{{47}^{o}}+\sin {{61}^{o}}-\sin {{11}^{o}}-\sin {{25}^{o}} is equal to

A

sin⁡ 7o\sin \,{{7}^{o}}

B

cos⁡ 7o\cos \,{{7}^{o}}

C

sin⁡ 36o\sin \,{{36}^{o}}

D

cos⁡ 36o\cos \,{{36}^{o}}

Answer

cos⁡ 7o\cos \,{{7}^{o}}

Explanation

Solution

sin⁡47o−sin25o+sin⁡61o−sin⁡11o\sin {{47}^{o}}-sin{{25}^{o}}+\sin {{61}^{o}}-\sin {{11}^{o}} =2  cos⁡ 36o sin⁡11o+2cos⁡ 36o sin⁡25o=2\,\,\cos \,{{36}^{o}}\,\sin {{11}^{o}}+2\cos \,{{36}^{o}}\,\sin {{25}^{o}} =2  cos⁡ 36o [sin⁡11o+ sin⁡25o]=2\,\,\cos \,{{36}^{o}}\,[\sin {{11}^{o}}+\,\sin {{25}^{o}}] =2  cos⁡ 36o[2 sin⁡ (25o+11o2) cos⁡ (25o−11o2)]=2\,\,\cos \,{{36}^{o}}\left[ 2\,\sin \,\left( \frac{{{25}^{o}}+{{11}^{o}}}{2} \right)\,\cos \,\left( \frac{{{25}^{o}}-{{11}^{o}}}{2} \right) \right] =4 cos⁡ 36o sin⁡18o cos⁡7o=4\,\cos \,{{36}^{o}}\,\sin {{18}^{o}}\,\cos {{7}^{o}} =4(5+14) (5−14) cos⁡ 7o=4\left( \frac{\sqrt{5}+1}{4} \right)\,\left( \frac{\sqrt{5}-1}{4} \right)\,\cos \,{{7}^{o}} =5−14 cos⁡ 7o=\frac{5-1}{4}\,\cos \,{{7}^{o}} = cos⁡ 7o=\,\cos \,{{7}^{o}}