The value of cos 52^circ + cos 68^circ + cos | vedclass

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(A) We use the sum-to-product formula: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$.Given expression: $\c

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(A) We use the sum-to-product formula: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$.Given expression: $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$.Applying the formula to the first two terms:$\cos 52^\circ + \cos 68^\circ = 2 \cos \left( \frac{52^\circ + 68^\circ}{2} ight) \cos \left( \frac{52^\circ - 68^\circ}{2} ight)$$= 2 \cos(60^\circ) \cos(-8^\circ)$$= 2 	imes \frac{1}{2} 	imes \cos(8^\circ) = \cos 8^\circ$.Now,the expression becomes: $\cos 8^\circ + \cos 172^\circ$.Using the formula again:$\cos 8^\circ + \cos 172^\circ = 2 \cos \left( \frac{8^\circ + 172^\circ}{2} ight) \cos \left( \frac{172^\circ - 8^\circ}{2} ight)$$= 2 \cos(90^\circ) \cos(82^\circ)$.Since $\cos 90^\circ = 0$,the entire expression equals $2 	imes 0 	imes \cos 82^\circ = 0$.

The value of $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$ is

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