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They are:

\begin{align} \cos(a)\cos(b)&=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big) \\[2ex] \sin(a)\sin(b)&=\frac{1}{2}\Big(\cos(a-b)-\cos(a+b)\Big) \\[2ex] \sin(a)\cos(b)&=\frac{1}{2}\Big(\sin(a+b)+\sin(a-b)\Big) \\[2ex] \cos(a)\sin(b)&=\frac{1}{2}\Big(\sin(a+b)-\sin(a-b)\Big) \\[2ex] \cos(a)+\cos(b)&=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex] \cos(a)-\cos(b)&=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \\[3ex] \sin(a)+\sin(b)&=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex] \sin(a)-\sin(b)&=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \end{align}

I have found nice mnemonics that helped me to remember the reduction formulae and others but I can't find a simple relationship between the formulas above. Can you help?

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  • $\begingroup$ What type of "relations" do you mean, or do you need ? $\endgroup$ – Jean Marie May 10 '16 at 8:33
  • $\begingroup$ 'Mathematics' is the subject which comes by practice! So use these formulas in questions and all will be stored in memory ;P $\endgroup$ – user5954246 May 10 '16 at 8:55
  • $\begingroup$ You are welcome to have a look at my answer to this post $\endgroup$ – Mick May 10 '16 at 16:57
  • $\begingroup$ @RichardSmith: Here's a diagram that may help. $\endgroup$ – Blue May 10 '16 at 17:12
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The only ones you need to know are the classical $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$ and $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. The others are mere consequences of those.

For example, by changing the signs, you get $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$. By summing, you have $\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)$, which is your first formula.

Similarly, by solving $p=a+b$ and $q=a-b$, you get the formula $\cos(p)+\cos(q) = 2\cos\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)$.

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  • 3
    $\begingroup$ And the two you suggest to remember can be derived in less than $30$ seconds using $e^{i(a+b)}=e^{ia}e^{ib}$ and Euler's formula :) $\endgroup$ – Guest Nov 20 '16 at 19:32
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Right away, you can cross off the fourth formula, since it is equivalent to the third formula after switching $a$ and $b$.

Then, you can also avoid the last four formulas, since these are all covered by the first three formulas via the relationships $$a+b = u, \quad a-b = v, \quad a = \frac{u+v}{2}, \quad b = \frac{u-v}{2}.$$

So that really leaves us with only three formulas. The first two are merely consequences of the cosine angle addition identity $$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b,$$ where a suitable addition or subtraction of the two forms of this equation are done; e.g., $$\begin{align*} \cos (a-b) &= \cos a \cos b + \sin a \sin b \\ \cos (a+b) &= \cos a \cos b - \sin a \sin b \\ \hline \cos(a-b) + \cos(a-b) &= 2 \cos a \cos b . \end{align*}$$ A similar concept applied to the sine angle addition identity yields the third (and fourth).

Of course, you can memorize the formulas, or re-derive them, but clearly it's faster to have more formulas memorized as long as you can remember them. What is important to stress is that a vast array of trigonometric identities are all consequences of some very basic identities, and these basic identities are the ones you really need to know.

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How about just restating the LHS. For example, you could restate $\cos a\sin b$ as $$\frac{\sin a\cos b +\cos a\sin b + \cos a\sin b - \sin a\cos b}{2}$$ and just figure it out from there. For Example, Let's start off with $\cos a\sin b$ and try to derive $\frac{1}{2}[\sin(a+b) - \sin(a-b)]$ $$ \begin{align} \cos a \sin b &= \frac{1}{2}\bigg[2\cos a\sin b\bigg] \\ &= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b\bigg] \\ &= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + 0\bigg] \\ &= \color{red}{\frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + (\sin a \cos b - \sin a\cos b)\bigg]} \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) +(\cos a\sin b - \sin a\cos b)\bigg] \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos b - \cos a\sin b )\bigg] \\ &= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos (-b) + \cos a\sin (-b) )\bigg] \\ &=\frac{1}{2}\bigg[\sin(a+b) - \sin(a-b)\bigg] \end{align} $$

Usually I just remember/figure out the red line.

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