My opinion is that there is no need to remember the formula for sine angle addition, that is, that $\sin(a)\cos(b)+\sin(b)\cos(a) = \sin(a+b)$. My reasoning is as follows:

Case 1: If you have both angles $a$ and $b$, simply calculate $\sin(a+b)$ directly.

Case 2: Let's say you start off with, not the angles, but the actual measurements themselves, to which the sine and cosine of a will be $y_1$ and $x_1$, and the sine and cosine of $b$ will be $y_2$ and $x_2$. From here, we can calculate angles $a$ and $b$ via $\arctan(y_1/x_1)$ and $\arctan(y_2/x_2)$, respectively, and then just do what we did previously in case 1.

Additionally, if you have one angle, but you only have the $x/y$ measurements of the other angle, then this is really just case 2 again.

Thus, the sine angle addition formula isn't necessary in order to advance ones knowledge of trigonometry, and can be safely done away with (if my reasoning is correct).

With this, is it safe to assume that I can file the sine angle addition formula under the "nice to know" category? Or is there something right under my nose which I've blindly missed? Honestly, this could very well be the case, hence the reason for this post to begin with.

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    $\begingroup$ Counterargument: how would you propose to compute the derivative of $\sin x$ by definition? $\endgroup$ Commented Jun 4, 2023 at 0:13
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    $\begingroup$ Case 3: you need to perform symbolic manipulations e.g. prove the identity $\,\sin 2a = 2 \sin a \cos a\,$. $\endgroup$
    – dxiv
    Commented Jun 4, 2023 at 0:13
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    $\begingroup$ Its useful for all sorts of integral identities. $\endgroup$ Commented Jun 4, 2023 at 0:13
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    $\begingroup$ To advance your knowledge it is good to remember it. To advance your understanding, it is even better to be able to derive it $\endgroup$
    – Amit
    Commented Jun 4, 2023 at 0:15
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    $\begingroup$ It is an important identity that you study in your trig course but actually learn through repetitive application because it comes up a lot. For example, we don't need to learn half angle and double angle formulas if we know the sine and cosine of sums. You can derive those formulas using geometry but they become easy to derive once you've been exposed to Euler's identity for complex numbers. $\endgroup$
    – John Douma
    Commented Jun 4, 2023 at 0:22

1 Answer 1


Suppose you have a weight, tied to a spring, bouncing up and down (technical term: simple harmonic motion). If you set up your coordinates right, the displacement (that is, the position) of the weight at time $t$ will be $\sin t$. Suppose you want to know how fast it's moving at time $t$. If you're willing to settle for its average velocity between time $t$ and a little bit later, time $t+h$ for a small number $h$, then velocity is distance travelled divided by length of time to do the travelling, which is $$ (\sin(t+h)-\sin t)/h $$ But if you want to know the velocity exactly at time $t$, then you have to know what happens to that expression as the time interval $h$ goes to zero. So, you use the addition formula to go $$ {\sin(t+h)-\sin t\over h}={\sin t\cos h+\cos t\sin h-\sin t\over h}={\cos h-1\over h}\sin t+{\sin h\over h}\cos t $$ Then with a little bit of geometry you can prove that as $h$ goes to zero, $(\cos h-1)/h$ goes to zero, and $(\sin h)/h$ goes to one, so the instantaneous velocity of the weight at time $t$ is $\cos t$.

This is what @Sean Roberson had in mind, though I have avoided using the word derivative.

  • $\begingroup$ Thank you very much for taking the time to explain it to me, this will be very useful :) $\endgroup$ Commented Jun 7, 2023 at 6:46

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