Question

Trigonometrie clasa a 9 -a

Answer 1

Răspuns............................................

Explicație pas cu pas:

Answer 2

$\it 2sin\left(\dfrac{\pi}{4}+x\right)sin\left(x-\dfrac{\pi}{4}\right)=2(sin\dfrac{\pi}{4}cos x+sin xcos\dfrac{\pi}{4})(sin xcos\dfrac{\pi}{4}-sin\dfrac{\pi}{4}cos x)=\\ \\ = 2\left(\dfrac{\sqrt2}{2}cos x+\dfrac{\sqrt2}{2}sinx\right)\left(\dfrac{\sqrt2}{2}sinx-\dfrac{\sqrt2}{2}cosx\right)=\\ \\ \\=2\cdot\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt2}{2}(sinx+cosx)(sinx-cosx)= sin^2x-cos^2x$