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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 $0\le A,B \le 90^{\circ}$.

Américo Tavares
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brianmearns
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    See this: http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities 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. – Michael Hardy Sep 28 '12 at 13:59
  • @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! – brianmearns Sep 28 '12 at 14:11
  • You cannot hope for an unambiguous solution, because if you know $\cos A=\cos B=0$ that is not enough to fix $A$ and $B$ well enough to get a handle on their average. $\cos\frac{A+B}2$ could be either $-1$, $0$ or $1$. – hmakholm left over Monica Sep 28 '12 at 15:02
  • @HenningMakholm: Sorry, I forgot to specify that both angles are first quadrant: 0 <= A, B <= 90-degrees. I'll edit the question. – brianmearns Sep 28 '12 at 15:26
  • Why are you expecting a simpler "solution"? And what do you want to use it for? – Mark Bennet Sep 28 '12 at 15:41
  • @MarkBennet: I'm not necessarily expecting a simpler answer, just hoping for one. I'm trying to work out an iterative approach to refining an estimation of the sine of an angle which can be done relatively quickly by hand. The slope of the secant that connects the lower reference angle (A) to x (the point in question) is somewhere between cos(A) and cos(B), so if I can find the cos of a particular angle between A and B, I can narrow down the interval and repeat, until I have two points close to x, and their cosines. – brianmearns Sep 28 '12 at 15:52
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    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}

     – Américo Tavares Sep 28 '12 at 16:47
  • @AméricoTavares: Well I can't see how either, that's why I asked =). It's not that it's ugly, it's that it's hard to compute by hand. I don't mind doing one or two 3 or 4 digit multiplications by hand, but finding the square root of numbers in [0, 1] is not really practical. – brianmearns Sep 28 '12 at 17:03
  • I think that the numerical computation by hand can only be carried out in a few particular cases. – Américo Tavares Sep 28 '12 at 18:13
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    @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. – robjohn Sep 28 '12 at 22:45
  • @robjohn: Thanks, that's a good explanation for what seems to be the running theme in most of these responses =). I'll concede that there's not likely a better solution that doesn't involve square roots. – brianmearns Oct 01 '12 at 12:25
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    @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. – brianmearns Oct 01 '12 at 12:30
  • @bmearns Glad to help. – Américo Tavares Oct 01 '12 at 15:40

2 Answers2

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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, $$ \cos\left(\frac{A+B}2\right)=\frac12\sqrt{(1+\cos A)(1+\cos B)}-\frac12\sqrt{(1-\cos A)(1-\cos B)}. $$

Did
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  • But OP says: "I know the cosines of A and B". So one still has to compute $\sin A$ and $\sin B$. – Américo Tavares Sep 28 '12 at 22:01
  • @AméricoTavares Do you really see this as a problem in the formula of my post? – Did Sep 29 '12 at 06:03
  • 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. – Américo Tavares Sep 29 '12 at 10:36
  • 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. – brianmearns Oct 01 '12 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.

Berci
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  • And how do you compute $\cos\frac{\alpha-\beta}{2}$? – Américo Tavares Sep 28 '12 at 16:51
  • 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. – brianmearns Sep 28 '12 at 17:08
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    then take the other solution – Berci Sep 29 '12 at 11:12