The value of cos^2 frac{pi}{12} + cos^2 frac{ | vedclass

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Question

(A) We need to evaluate the expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$.Using the identity $\cos^2 	heta = \f

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$\frac{3}{2}$

Vedclass · Answer

(A) We need to evaluate the expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$.Using the identity $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}$,we can rewrite the expression as:$= \frac{1 + \cos(\frac{\pi}{6})}{2} + \cos^2(\frac{\pi}{4}) + \frac{1 + \cos(\frac{5\pi}{6})}{2}$$= \frac{1}{2} + \frac{1}{2} \cos(\frac{\pi}{6}) + (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} + \frac{1}{2} \cos(\frac{5\pi}{6})$$= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} (\cos(\frac{\pi}{6}) + \cos(\frac{5\pi}{6}))$$= \frac{3}{2} + \frac{1}{2} (\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})$$= \frac{3}{2} + 0 = \frac{3}{2}$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

The value of $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$ is

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