# How do I derive the identity $\sin(a+b)=\sin(a)\cos(b) + \cos(a)\sin(b)$ using the Unit Circle

I'd like it explained through the unit circle as I find trig identities easier much easier to understand in this manner.

EDIT: I know you have to apply the identity $\sin(x)=\cos(90-x)$, but I'm wondering how i'd visualise all this on the unit circle?

## 5 Answers

This is the proof of $\cos(a+b)=\cos a \, \cos b - \sin a \, \sin b$.

See also this proof.

Using complex numbers and exponential form perhaps help (at least algebraically) to digest these trigonometric addition formulas:

All we have to know is $\cos a+i\cdot\sin a=e^{ai}$ for any $a\in\Bbb R$, and that $i^2=-1$, and that $e^{x+y}=e^x\cdot e^y$ for any $x,y\in\Bbb C$. Then calculate both sides of $e^{(a+b)i}=e^{ai}e^{bi}$.

If you prefer, instead, you can use the matrices of rotation: $$R_a:=\pmatrix{\cos a&-\sin a\\ \sin a &\cos a}$$ and use matrix multiplication to verify the identities, knowing that $$R_{a+b}=R_a\cdot R_b \ .$$

• I'm afraid I haven't learned some of the concepts mentioned in your answer, thank you for this however. – seeker Mar 3 '13 at 15:31
• And how do you know all those properties about complex numbers? – André Caldas Mar 3 '13 at 16:03
• Any proofs I know of for Euler's identity require the knowledge that $\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx} \cos x = - \sin x$. The limit involved in these derivatives requires knowledge of the trig subtraction formulas, which means that the proof using this identity (as far as I can tell) is circular. – Omnomnomnom Jan 14 '14 at 22:49
• The rotation matrix proof, however, isn't a bad idea. I have a hunch that it is essentially the same as the usual geometric proof. – Omnomnomnom Jan 14 '14 at 22:52

This video will clear matters beautifully. Make sure to watch it, and then the next.

The explanation given by eepsmedia here is what you are looking for. Although his argument gives the cosine addition formula, and only in the case when $\alpha+\beta < \pi/2$, you should be able to use the same methods to obtain the sine angle addition formula.

If the reader knows a little calculation, we have an elegant proof of this identity. Let $$f(x) = \sin(x + b) - \sin x\cos b - \sin b\cos x \ \Rightarrow$$ $$f^{\prime}(x) = \cos(x + b) - \cos x \cos b + \sin b \sin x = \cos(x + b) - \cos(x + b) = 0$$ since $\cos(x + b) = \cos x \cos b - \sin b \sin x$. Thus, $$f(x) = \sin(x + b) - \sin x\cos b - \sin b\cos x = C, \quad C \in \mathbb{R}$$ For $x = 0$, we have $C = 0$. In particular, at $x = a$, $$\boxed{\sin(a + b) = \sin a\cos b + \sin b\cos a}$$