cos frac{7 pi}{8}+cos frac{pi}{4}+cos left(fr | vedclass

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(C) Given expression: $\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}ight)-1$. Using $\cos(-	heta) = \cos 	heta$,the expression

Vedclass · Accepted Answer

$4 \cos \frac{\pi}{16} \cos \frac{3 \pi}{8} \cos \frac{9 \pi}{16}$

Vedclass · Answer

(C) Given expression: $\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}ight)-1$. Using $\cos(-	heta) = \cos 	heta$,the expression becomes $\cos \frac{7 \pi}{8} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{8} - 1$. Since $\cos \frac{7 \pi}{8} = \cos(\pi - \frac{\pi}{8}) = -\cos \frac{\pi}{8}$,the expression simplifies to: $-\cos \frac{\pi}{8} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{8} - 1 = \frac{1}{\sqrt{2}} - 1$. Now,evaluating option $(C)$: $4 \cos \frac{\pi}{16} \cos \frac{3 \pi}{8} \cos \frac{9 \pi}{16}$. $= 2 \left(2 \cos \frac{9 \pi}{16} \cos \frac{\pi}{16}ight) \cos \frac{3 \pi}{8}$. Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $= 2 \left(\cos \frac{10 \pi}{16} + \cos \frac{8 \pi}{16}ight) \cos \frac{3 \pi}{8} = 2 \left(\cos \frac{5 \pi}{8} + \cos \frac{\pi}{2}ight) \cos \frac{3 \pi}{8}$. Since $\cos \frac{\pi}{2} = 0$,this becomes $2 \cos \frac{5 \pi}{8} \cos \frac{3 \pi}{8}$. $= \cos(\frac{5 \pi}{8} + \frac{3 \pi}{8}) + \cos(\frac{5 \pi}{8} - \frac{3 \pi}{8}) = \cos \pi + \cos \frac{\pi}{4} = -1 + \frac{1}{\sqrt{2}}$. Thus,option $(C)$ is correct.

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

$\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}\right)-1=$

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