# application of fundamental theorem of integral calculus

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

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

These the addition formulas (http://planetmath.org/GoniometricFormulae)

 $\sin(x\!+\!\alpha)\;=\;\sin{x}\cos\alpha+\cos{x}\sin\alpha,$
 $\cos(x\!+\!\alpha)\;=\;\cos{x}\cos\alpha-\sin{x}\sin\alpha.$
Title application of fundamental theorem of integral calculus ApplicationOfFundamentalTheoremOfIntegralCalculus 2013-03-22 18:50:52 2013-03-22 18:50:52 pahio (2872) pahio (2872) 6 pahio (2872) Example msc 26A06 TrigonometricFormulasFromSeries