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?

  • $\begingroup$ 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

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 $\cos(p)+\cos(q) = 2\cos\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)$.

  • 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

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.


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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