The value of (1+cos frac{pi}{6})(1+cos frac{p | vedclass

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Question

(A) Given expression: $(1+\cos \frac{\pi}{6})(1+\cos \frac{\pi}{3})(1+\cos \frac{2\pi}{3})(1+\cos \frac{7\pi}{6})$Using the values of trigonometric fu

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(A) Given expression: $(1+\cos \frac{\pi}{6})(1+\cos \frac{\pi}{3})(1+\cos \frac{2\pi}{3})(1+\cos \frac{7\pi}{6})$Using the values of trigonometric functions:$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, $\cos \frac{\pi}{3} = \frac{1}{2}$, $\cos \frac{2\pi}{3} = -\frac{1}{2}$, $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$Substituting these values:$= (1 + \frac{\sqrt{3}}{2})(1 + \frac{1}{2})(1 - \frac{1}{2})(1 - \frac{\sqrt{3}}{2})$Group the terms using the identity $(a+b)(a-b) = a^2 - b^2$:$= [(1 + \frac{\sqrt{3}}{2})(1 - \frac{\sqrt{3}}{2})] 	imes [(1 + \frac{1}{2})(1 - \frac{1}{2})]$$= (1^2 - (\frac{\sqrt{3}}{2})^2) 	imes (1^2 - (\frac{1}{2})^2)$$= (1 - \frac{3}{4}) 	imes (1 - \frac{1}{4})$$= \frac{1}{4} 	imes \frac{3}{4} = \frac{3}{16}$

The value of $(1+\cos \frac{\pi}{6})(1+\cos \frac{\pi}{3})(1+\cos \frac{2\pi}{3})(1+\cos \frac{7\pi}{6})$ is

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