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Why does $r\cos{θ} \cos{a} - r\sin{θ}\sin{a} = r\cos(θ + a)$?

My book writes this but doesn't show the in-between steps and I can't remember trigonometry enough to see why this works.

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    It's a trig identity, you can find the proof of it on Wikipedia, it's just angle sum for cosine – 2017-01-22
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    That $r$ in front only complicates things. You want to show that $$\cos a\cos b-\sin a\sin b=\cos (a+b)$$ It is a known identity as @TehRod mentions. You can find the-geometric-solution easily on line. – 2017-01-22
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    $$\cos{(\alpha+\beta)}\equiv \cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}$$ This is a common identity. Here is a [geometric proof](http://www.intmath.com/analytic-trigonometry/2-sum-difference-angles.php) for each of $\sin(\alpha+\beta)$, $\cos(\alpha+\beta)$ and $\tan(\alpha+\beta)$. – 2017-01-22

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The answer is becuase you can factor out an $r$ and use the identity for cosine:

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

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A bit of an overkill here, I just post it as an "alternative" solution. The following hold true: $$\cos a=\frac{e^{ia}+e^{-ia}}{2}\\\cos \theta=\frac{e^{i\theta}+e^{-i\theta}}{2}\\\sin a= \frac{e^{ia}-e^{-ia}}{2i}\\ \sin \theta= \frac{e^{i\theta}-e^{-i\theta}}{2i}\\\cos (a+\theta)=\frac{e^{i(a+\theta)}+e^{-i(a+\theta)}}{2}$$

Calculate LHS $(\cos{θ} \cos{a} - \sin{θ}\sin{a})$ based on these and you get the result.

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For the positive acute angle case, try using the cosine law on this triangle. Nothing better for understanding than productive struggle.

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