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I am trying to evaluate $\cos(\alpha+\beta+\gamma)$

This is what I have done so far:

I know $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

and $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

Treating $\cos(\alpha+\beta+\gamma)$ as $\cos[(\alpha+\beta)+\gamma]$

means that I can write $\cos(\alpha+\beta+\gamma) = \cos(\alpha +\beta) \cos\gamma - \sin(\alpha+\beta)\sin\gamma$

Taking the $\cos(\alpha +\beta) \cos\gamma$ part first: $\cos(\alpha +\beta) \cos\gamma= \cos\alpha\cos\beta\cos\gamma -\sin\alpha\sin\beta\cos\gamma$

and here is the part where I am struggling with getting the signs correct:

$- \sin(\alpha+\beta)\sin\gamma = -\sin\alpha\cos\beta\sin\gamma - \cos\alpha\sin\beta\sin\gamma$

To give $\cos(\alpha+\beta+\gamma) = \cos\alpha\cos\beta\cos\gamma -\sin\alpha\sin\beta\cos\gamma-\sin\alpha\cos\beta\sin\gamma - \cos\alpha\sin\beta\sin\gamma$

I am really unsure that I have my signs correct.

dajoker
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mikoyan
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1 Answers1

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A hint in your formula's favor is that if we permute the variables, $\alpha,\beta,\gamma$, you get the same formula. Thus, it is necessary that if the coefficient of $\sin\alpha \sin\beta\cos \gamma$ is $-1$, then the other two terms with two $\sin$s in them would also have to have coefficients $-1$.

A general rule is, if you are worried about signs, check with specific examples.

What happens when $\beta=-\alpha$? Then your formula would be:

$$\cos\alpha\cos(-\alpha)\cos\gamma -\sin\alpha\sin(-\alpha)\cos\gamma-\sin\alpha\cos(-\alpha)\sin\gamma - \cos\alpha\sin(-\alpha)\sin\gamma =\\ \cos\gamma(\cos^2\alpha + \sin^2\alpha) - \sin\alpha\cos\alpha\sin\gamma + \cos\alpha\sin\alpha\sin\gamma\\ = \cos \gamma = \cos(\alpha+\beta+\gamma)$$

what happens when $\alpha=\beta=45^\circ$? then $\cos(\alpha+\beta+\gamma)=-\sin\gamma$.

Since $\sin\alpha=\sin\beta=\cos\alpha=\cos\beta=\sqrt{\frac{1}{2}}$, you can plug in and check again:

$$\frac{1}{2}\cos\gamma - \frac{1}{2}\cos\gamma -\frac{1}{2}\sin\gamma -\frac{1}{2}\sin\gamma = -\sin\gamma$$

When you get to complex numbers, you'll see an elegant way to "see" this is the right formula. The formula for $\cos(\alpha+\beta+\gamma+\dots)$ will in general have terms with an even number of $\sin$ expressions, and the coefficient for each term will be $1$ if there is a multiple of $4$ $\sin$ terms, and $-1$ if there are not.

Thomas Andrews
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