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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?

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This isn't on the unit circle, but I like it: – Blue Mar 3 '13 at 16:00

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

See also this proof.

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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 $$R_{a+b}=R_a\cdot R_b \ .$$

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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.

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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.

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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 $x = a$, $$ \boxed{\sin(a + b) = \sin a\cos b + \sin b\cos a} $$

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