Question

a) Cos 8x -Cos 2x

b) Provar que:
Sen 30graus + Sen 10graus/ COS 30 graus + Cos 10graus = Tg 20graus

Answer 1

$\mathrm{\mathbf{a)}\ \cos{a}-\cos{b}=-2\sin{\bigg(\dfrac{a+b}{2}\bigg)}\sin{\bigg(\dfrac{a-b}{2}\bigg)}\ \to}\\\\ \mathrm{\to\ \cos{8x}-\cos{2x}=-2\sin{\bigg(\dfrac{8x+2x}{2}\bigg)}\sin{\bigg(\dfrac{8x-2x}{2}\bigg)}=}\\\\ \mathrm{=-2\sin{\bigg(\dfrac{10x}{2}\bigg)}\sin{\bigg(\dfrac{6x}{2}\bigg)}=\boxed{\mathbf{-2\sin{5x}\sin{3x}}}}$

$\mathrm{\mathbf{b)}\ \sin{a}+\sin{b}=2\sin{\bigg(\dfrac{a+b}{2}\bigg)}\cos{\bigg(\dfrac{a-b}{2}\bigg)}}\\\\ \mathrm{\cos{a}+\cos{b}=2\cos{\bigg(\dfrac{a+b}{2}\bigg)}\cos{\bigg(\dfrac{a-b}{2}\bigg)}\ \to}\\\\ \mathrm{\to\ \dfrac{\sin{30\º}+\sin{10\º}}{\cos{30\º}+\cos{10\º}}=\dfrac{2\sin{\bigg(\dfrac{30+10}{2}\bigg)}\cos{\bigg(\dfrac{30-10}{2}\bigg)}}{2\cos{\bigg(\dfrac{30+10}{2}\bigg)}\cos{\bigg(\dfrac{30-10}{2}\bigg)}}=}\\\\ \mathrm{=\dfrac{2\sin{20\º}\cos{10\º}}{2\cos{20\º}\cos{10\º}}=\dfrac{\sin{20\º}}{\cos{20\º}}=\boxed{\mathbf{\tan{20\º}}}}$