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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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    $\cos B=2\cos^2\frac B2-1$. – 2017-01-22
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    But why can't we use cos1/2B if we need to find that? For cos2B we use cos2B=... not something else – 2017-01-22
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    You could, but you don't have a value for $\cos \frac B4$, so that doesn't help. Also please use parentheses. It is clearer to say cos(B/2) than cos 1/2B as the B could be in the denominaor. Better yet, learn to use MathJax as shown in the FAQ. – 2017-01-22

3 Answers 3

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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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    Yeah, I tried that one now, and that works, but why do we use the equation for cos B, if it asks for cos 1/2B? – 2017-01-22
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    @JohnFire : Because you're looking for an equation that lets you write $\cos \frac12 b$ in terms of $\cos b$. So if you have an equation containing both, you can solve it for one expression in terms of the other. – 2017-01-22
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Hint

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

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    Yeah, but if we use the formula for cos2x when we find cos2B, then why don't we use the formula for cos1/2x if we find cos1/2B, and why do we use cosx's formula? – 2017-01-22
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    But if you need $B/2$ why are you looking for $1/(2B)$? – 2017-01-22
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    Because B/2 = 1/2 B? Same thing – 2017-01-22
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    I think someone is confusing $1/(2b)$ and $(1/2)b$. Without parentheses, the in-line fraction notation is ambiguous. – 2017-01-22
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    Oh, I see: you mean $B/2=(1/2)B$. You can do that but see that you are going down with the angle. If you start with $B$ and want $B/2$ is not a good idea use the formula for $B/2$, because like you saw, you go down with the angle. – 2017-01-22
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    @Arnaldo Yeah, sorry i need to get used to the formatting. What do you mean by going down with the angle? – 2017-01-22
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    @JonhFire: I mean, if you use $B/2$ At the left side of the formula you get $B/4$ at the right site (the angle goes down). But you have $B$, That is why is better start with $B$ at the left side and get $B/2$. Is it clear? – 2017-01-22
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    @Arnaldo Oh, okay, I get it now, thank you :) – 2017-01-23
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    @JohnFire: you are very welcome – 2017-01-23
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$\cos 2B = 2 \cos^2 B - 1 = \frac9{8} - 1 = \frac{1}{8}$

Second answer -

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

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

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

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

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

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