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I got $\cos 2B$; $\cos 2B=2 \cos^2 B - 1 =9/8 -1 =1/8$

but when I tried cos1/2B, I got: COS1/2B=cos^2*1/4B-1, then I solved it, and didn't get the answer which was in the book, why?

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3 Answers

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Hint: Since $\cos 2b =\cos^2 b-\sin^2 b =2\cos^2 b -1$, we have $$\cos b= \cos 2(\tfrac12 b)=2\cos^2 \tfrac12b-1$$ $$\cos\tfrac12 b =\pm\sqrt{\tfrac12(1+\cos b)}$$

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Hint

$$\cos x= 2\cos^2(x/2)-1$$

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$\cos 2B = 2 \cos^2 B - 1 = \frac9{8} - 1 = \frac{1}{8}$

Second answer -

$\cos B = 2 \cos^2(\frac B2) - 1$

$\frac34 = 2 \cos^2(\frac B2) - 1$

$\frac 74 = 2 \cos^2(\frac B2)$

$\frac 78 = \cos^2(\frac B2)$

$\sqrt{\frac78} = \cos (\frac B2)$

$\frac12 \sqrt{\frac72} = \cos (\frac B2)$

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