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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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    $\begingroup$ See this: en.wikipedia.org/wiki/… Maybe what's more interesting is this expression for the tangent of an average: $\tan((\alpha+\beta)/2)= (\sin\alpha+\sin\beta)/(\cos\alpha+\cos\beta)$. That's form of the tangent half-angle formula. $\endgroup$ Sep 28, 2012 at 13:59
  • $\begingroup$ @MichaelHardy: Thanks for the lead, I've actually already spent some time looking at those identities but haven't found a way to apply them. Using the tangent of an average just turns the problem into finding the sine of the average, instead of the cosine. Do you actually know how to use these to solve my problem? Thanks! $\endgroup$ Sep 28, 2012 at 14:11
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    $\begingroup$ I cannot see how can you find something different from what you consider "ugly expressions": \begin{eqnarray*} \cos \frac{A+B}{2} &=&\cos \frac{A}{2}\cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2} \\ &=&\pm \frac{1}{2}\sqrt{\left( 1+\cos A\right) \left( 1+\cos B\right) } \\ &&\mp \frac{1}{2}\sqrt{\left( 1-\cos A\right) \left( 1-\cos B\right) }. \end{eqnarray*} $\endgroup$ Sep 28, 2012 at 16:47
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    $\begingroup$ @bmearns: Consider the case where $A=0^\circ$ and $B=90^\circ$. $\cos(A)=1$ and $\cos(B)=0$; however, $\cos\left(\frac{A+B}{2}\right)=\cos(45^\circ)=\frac{\sqrt{2}}{2}$. The cosines of $A$ and $B$ are about as nice as you could ask for, yet the cosine of $\frac{A+B}{2}$ still involves a square root. $\endgroup$
    – robjohn
    Sep 28, 2012 at 22:45
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    $\begingroup$ @AméricoTavares: I actually liked you answer better than the accepted one: sign ambiguity is not an issue since all of my angles are first quadrant. Thanks for the help. $\endgroup$ Oct 1, 2012 at 12:30

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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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  • $\begingroup$ But OP says: "I know the cosines of A and B". So one still has to compute $\sin A$ and $\sin B$. $\endgroup$ Sep 28, 2012 at 22:01
  • $\begingroup$ @AméricoTavares Do you really see this as a problem in the formula of my post? $\endgroup$
    – Did
    Sep 29, 2012 at 6:03
  • $\begingroup$ No, I don't. Of course you did know how to express $\sin A$ and $\sin B$ in terms of $\cos A$ and $\cos B$. Your last formula has no sign ambiguity, which is an advantage over mine, now deleted. $\endgroup$ Sep 29, 2012 at 10:36
  • $\begingroup$ 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. $\endgroup$ Oct 1, 2012 at 12:29
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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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  • $\begingroup$ And how do you compute $\cos\frac{\alpha-\beta}{2}$? $\endgroup$ Sep 28, 2012 at 16:51
  • $\begingroup$ Agreed. This is a nice attempt, but you've changed the problem from finding the cosine of the average to finding the cosine of half the difference. The point is really that I only the cosines (and sines) for A and B, and need to find the cosine of their average without relying on the sines or cosines or any other angles. $\endgroup$ Sep 28, 2012 at 17:08
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    $\begingroup$ then take the other solution $\endgroup$
    – Berci
    Sep 29, 2012 at 11:12

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