Question

Arătați că are loc egalitatea din poză​

Answer 1

Răspuns:

Formule folosite: $sin(x+y)*sin(x-y)=sin^2x-sin^2y$

$sin(x+y)*sin(x-y)=cos^2y-cos^2x$

$\frac{cos^2(5x)-cos^2(4x)}{sin^2(7x)-sin^2(2x)}=-\frac{sinx}{sin(5x)}$

$\frac{sin(4x+5x)*sin(4x-5x)}{sin(7x+2x)*sin(7x-2x)}=-\frac{sinx}{sin(5x)}$

$\frac{sin(9x)*sin(-x)}{sin(9x)*sin(5x)}=-\frac{sinx}{sin(5x)}$

$-\frac{sinx}{sin(5x)}=-\frac{sin}{sin(5x)}$ "Adevarat

Answer 2

Folosim formulele trigonometrice de transformarea sumelor

în produse.

$\it cos^2a-cos^2b=(cosa-cosb)(cosa+cosb)=\\ \\ \\ -2sin\dfrac{a+b}{2}sin\dfrac{a-b}{2}\cdot2cos\dfrac{a+b}{2}cos\dfrac{a-b}{2}=\\ \\ \\ =(-2sin\dfrac{a-b}{2}\cdot cos\dfrac{a-b}{2})\cdot(2sin\dfrac{a+b}{2}\cdot cos\dfrac{a+b}{2})=-sin(a-b)sin(a+b)$

$\it {Pentru\ \ a=5x,\ \ b=4x,\ \ vom\ \ avea:}\\ \\ \it cos^25x-cos^24x=-sin(5x-4x)\cdot sin(5x+4x)=-sinx\cdot sin9x\ \ \ \ \ (1)$

$\it sin^2a-sin^2b=(sina-sinb)(sina+sinb)=\\ \\ \\ = 2sin\dfrac{a-b}{2}cos\dfrac{a+b}{2}\cdot 2sin\dfrac{a+b}{2}cos\dfrac{a-b}{2}=\\ \\ \\ =(2sin\dfrac{a-b}{2}cos\dfrac{a-b}{2})\cdot (2sin\dfrac{a+b}{2}cos\dfrac{a+b}{2})=sin(a-b)\cdot sin(a+b)$

$\it Pentru\ \ a=7x,\ \ b=2x,\ \ vom\ \ avea:\\ \\ sin^27x-sin^22x=sin(7x-2x)\cdot sin(7x+2x)=sin5x\cdot sin9x\ \ \ \ \ \ (2)$

$\it (1),\ (2)\ \Rightarrow\ \dfrac{cos^25x-cos^24x}{sin^27x-sin^22x}=\dfrac{-sinx\cdot sin9x}{sin5x\cdot sin9x}=\dfrac{-sinx}{sin5x}=-\dfrac{sinx}{sin5x}$