[tex]P=sin\left(\dfrac{x}2\right)\cdot sin\left(\dfrac{5x}2\right)=sin\left(\dfrac{5x}2\right)\cdot sin\left(\dfrac{x}2\right)=\dfrac{1}2\cdot\left[cos\left(\dfrac{5x}2-\dfrac{x}2\right)-cos\left(\dfrac{5x}2+\dfrac{x}2\right)\right]=\\\\=\dfrac{1}2\cdot[cos(2x)-cos(3x)]=\dfrac{1}2[2cos^2x-1-cos(3x)].\\\\cos(3x)=cos(2x+x)=cos(2x)\cdot cosx-sin(2x)\cdot sinx=\\=(2cos^2x-1)\cdot cosx-2sin^2x\cdot cosx=2cos^3x-cosx-2(1-cos^2x)cosx=\\=2cos^3x-cosx-2cosx+2cos^3x=4cos^3x-3cosx=4a^3-3a.\\\\Deci\ P=\dfrac{1}2(2a^2-1-4a^3+3a)=\dfrac{-4a^3+2a^2+3a-1}2.[/tex]
Învață formulele de la trigonometrie și fă multe exerciții !
Green eyes.
