Sum and Difference Formulas for Sine and Cosine

Theorem. For all real $x, y$,

Proof. Let $P(\cos x,\sin x)$, $Q(\cos y,\sin y)$, $P'(\cos(x-y),\sin(x-y))$ and $A(1,0)$

all be points on the unit circle, corresponding to the angles $x$, $y$, $x-y$ and $0$

respectively. The angle from $OQ$ to $OP$ is $x-y$, and the angle from $OA$ to $OP'$ is also

$x-y$; since equal angles at the centre of a circle of a given radius cut off chords of equal

length, chord $PQ$ = chord $P'A$, i.e. $PQ^2=P'A^2$.

Compute each side using the distance formula. First,

Grouping and using the fundamental identity ($\sin^2x+\cos^2x=1$, and the same for $y$):

Next,

again using $\sin^2(x-y)+\cos^2(x-y)=1$. Setting $PQ^2=P'A^2$:

Deducing $\cos(x+y)$. Write $x+y=x-(-y)$ and apply the theorem just proved with $y$

replaced by $-y$:

Since cosine is an even function ($\cos(-y)=\cos y$) and sine is odd ($\sin(-y)=-\sin y$),

Together, $\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$.

Deducing $\sin(x+y)$. Using the co-function identity $\sin\theta=\cos\left(\tfrac{\pi}{2}-\theta\right)$

(itself an immediate case of $\cos(x-y)$ with $x=\pi/2,\ y=\theta$: $\cos(\pi/2-\theta)= \cos\tfrac\pi2\cos\theta+\sin\tfrac\pi2\sin\theta=0+\sin\theta$),

Applying the $\cos(x-y)$ theorem with $\tfrac\pi2-x$ in place of $x$:

using $\cos(\pi/2-x)=\sin x$ and $\sin(\pi/2-x)=\cos x$. …