Folosim formulele trigonometrice de transformarea sumelor
în produse.
[tex]\it cos^2a-cos^2b=(cosa-cosb)(cosa+cosb)=\\ \\ \\ -2sin\dfrac{a+b}{2}sin\dfrac{a-b}{2}\cdot2cos\dfrac{a+b}{2}cos\dfrac{a-b}{2}=\\ \\ \\ =(-2sin\dfrac{a-b}{2}\cdot cos\dfrac{a-b}{2})\cdot(2sin\dfrac{a+b}{2}\cdot cos\dfrac{a+b}{2})=-sin(a-b)sin(a+b)[/tex]
[tex]\it {Pentru\ \ a=5x,\ \ b=4x,\ \ vom\ \ avea:}\\ \\ \it cos^25x-cos^24x=-sin(5x-4x)\cdot sin(5x+4x)=-sinx\cdot sin9x\ \ \ \ \ (1)[/tex]
[tex]\it sin^2a-sin^2b=(sina-sinb)(sina+sinb)=\\ \\ \\ = 2sin\dfrac{a-b}{2}cos\dfrac{a+b}{2}\cdot 2sin\dfrac{a+b}{2}cos\dfrac{a-b}{2}=\\ \\ \\ =(2sin\dfrac{a-b}{2}cos\dfrac{a-b}{2})\cdot (2sin\dfrac{a+b}{2}cos\dfrac{a+b}{2})=sin(a-b)\cdot sin(a+b)[/tex]
[tex]\it Pentru\ \ a=7x,\ \ b=2x,\ \ vom\ \ avea:\\ \\ sin^27x-sin^22x=sin(7x-2x)\cdot sin(7x+2x)=sin5x\cdot sin9x\ \ \ \ \ \ (2)[/tex]
[tex]\it (1),\ (2)\ \Rightarrow\ \dfrac{cos^25x-cos^24x}{sin^27x-sin^22x}=\dfrac{-sinx\cdot sin9x}{sin5x\cdot sin9x}=\dfrac{-sinx}{sin5x}=-\dfrac{sinx}{sin5x}[/tex]
