# application of fundamental theorem of integral calculus

We will derive the addition formulas of the sine and the cosine functions supposing known only their derivatives and the chain rule.

Define the function$F:\,\mathbb{R}\to\mathbb{R}$  through

 $F(x)\;:=\;[\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha)]^{2}+[\cos{x% }\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)]^{2}$

where $\alpha$ is, for the , a constant.  The derivative of $F$ is easily calculated:
$F^{\prime}(x)\;=\;\\ \mbox{\;\;}2[\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha)][\cos{x}% \cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)]\\ +2[\cos{x}\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)][-\sin{x}\cos\alpha-% \cos{x}\sin\alpha+\sin(x\!+\!\alpha)]$

But this expression is identically 0.  By the fundamental theorem of integral calculus, $F$ must be a constant function.  Since  $F(0)=0$,  we have

 $F(x)\;\equiv\;0$

for any $x$ and naturally also for any $\alpha$.  Because $F(x)$ is a sum of two squares, the both addends of it have to vanish identically, which yields the equalities

 $\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha)\;=\;0,\qquad\cos{x}\cos% \alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)\;=\;0.$

 $\sin(x\!+\!\alpha)\;=\;\sin{x}\cos\alpha+\cos{x}\sin\alpha,$
 $\cos(x\!+\!\alpha)\;=\;\cos{x}\cos\alpha-\sin{x}\sin\alpha.$