Proof of the angle sum identity for $\sin$ $$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$
How can I prove this statement?
 A: I typed "Proof of Trigonometric formulae" in to Google and the second hit was an extensive Wikipedia article which supplies proofs to many, many trigonometric identities.
Click here for the section that you want.
A: This is the way I learned it: a geometric proof like this on Wikipedia.
The segment $OP$ has length $1$. We have then, $\sin(\alpha + \beta) = PB = PR + RB = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)$.
A: $e^{i(a+b)}=\cos(a+b)+i\sin(a+b)$ by Euler's formula. But $e^{i(a+b)}=e^{ia}e^{ib}=(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))= \cos(a)\cos(b)-\sin(a)\sin(b)+i(\sin(a)\cos(b) + \cos(a)\sin(b))$
So by comparing real and imaginary parts you obtain the trigonometric addition formulae for both $\sin$ and $\cos$.
A: This gem comes from E. Schmidt: Consider $f(x)=\sin(\alpha+\beta-x)\cos(x)+\cos(\alpha+\beta-x)\sin(x)$. Since $f'(x)=0$ we know that $f$ is constant, hence $f(0)=f(\beta)$.
A: Here is a modern proof using sleight of hand.(1)An isometry of the Euclidean plane $E$ is a function $f:E\to E$ that preserves the distance $d$ between points: $d(A,B)=d(f(A),f(B))$ for all $A,B\in E$. We have (2):An isometry $f$ is onto.Because, for $Q\in E$, let $A_1,A_2,A_3\in E$ be non-co-linear with $Q\not \in \{f(A_1),f(A_2),f(A_3)\}$.The unique $P$ such that $d(P,A_i)=d(Q,f(A_i))$ for $i=1,2,3$ must satisfy $f(P)=Q$.We have (3):An isometry that fixes 3 non-co-linear points $A_1,A_2,A_3$ is the identity function.Because,as in (2),for any $P$, the only $Q$ that can satisfy  $ d(P,A_i)=d(Q,f(A_i))=d(Q,A_i)$ for $i=1,2,3$ is $Q=P$.We have (4):If isometries $ f,g$ agree on 3 non-co-linear points then $f=g$. Because $g^{-1}:E\to E$ exists by (2), and hence is also an isometry (obviously), and by (3) we have  $g^{-1}(f(P))=P$ for all $ P$, so $$g(P)=g(g^{-1}(f(P))=f(P).$$  Now to use all this :(5): Choose a pair of orthogonal co-ordinate axes for $E$ with origin $O$. Consider two types of isometry :First, $R_b$ is a counter-clockwise rotation about $O$ through angle $b$. Second, $M_b$ sends the point with co-ordinates $(x,y)$ to $$(x\cos b-y\sin b,x\sin b+y\cos b).$$The fact that $M_b$ is an isometry relies only on the theorem of Pythagoras, or in other words, $\cos^2 b+\sin^2 b=1$. Now for any $b$, observe that $R_b$ and $M_b$ agree on the 3 non-co-linear points $(0,0),(1,0),(0,1)$.Therefore by (4) we have $$R_b=M_b\text { for all b }.$$Finally, since $R_b$ and $R_c$ are rotations about $O$ we have$$R_bR_c=R_{b+c} \text{ for all b,c }.$$ Therefore we have $$M_{b+c}=R_{b+c}=R_bR_c=M_bM_c\text{ for all b,c}.$$ By comparing the co-ordinates of  $M_{b+c}(1,0)$ with the co-ordinates of $M_bM_c(1,0)$, we have the angle-sum formulas for $\cos$ and $\sin$.
