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Is there an expression for the cosine of an average of two angles? I.e., If I know the cosines of $A$ and $B$, can I easily find the cosine of $(A+B)/2$? Ideally, I'm looking for something that can be computed pretty easily by hand, for instance using addition, subtraction, and multiplication (division if really necessary). So far, I've only been able to use the rules for half angles and sum of angles to come up with two ugly expressions involving square roots, which are out of the question.

To clarify, both angles $A$ and $B$ are in the first quadrant, so $0\le A,B \le 90^{\circ}$.

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    @bmearns Glad to help.2012-10-01

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Since $A$ and $B$ are in $[0,\frac\pi2]$ and, for every $\theta$, $\cos(2\theta)=2\cos^2\theta-1$, $ \cos\left(\frac{A+B}2\right)=\sqrt{\frac{1+\cos(A+B)}2}=\sqrt{\frac{1+\cos A\cos B-\sin A\sin B}2}. $ In terms of $\cos A$ and $\cos B$ only, $ \cos\left(\frac{A+B}2\right)=\sqrt{\frac{1+\cos A\cos B-\sqrt{(1-\cos^2A)(1-\cos^2B)}}2}. $ This is equivalent to a formula indicated by @AméricoTavares in a comment, namely, $ \cos\left(\frac{A+B}2\right)=\frac12\sqrt{(1+\cos A)(1+\cos B)}-\frac12\sqrt{(1-\cos A)(1-\cos B)}. $

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    Thanks, @did. I'll concede that I can't escape using square roots for this, so I guess I'll have to practice Newton's method.2012-10-01
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Well, you can get $\cos\frac{\alpha+\beta}2 = \frac{\cos\alpha + \cos\beta}{2\color{grey}{\cos\frac{\alpha-\beta}2}}$ it can be said nice enough, much nicer is not likely.

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    then take the other solution2012-09-29