Question

Exercitiile 5, 9 si 10.

Answer 1

Salut,

Exerciţiul 10:

$cos\dfrac{\pi}{12}=cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=cos\dfrac{\pi}{3}\cdot cos\dfrac{\pi}{4}+sin\dfrac{\pi}{3}\cdot sin\dfrac{\pi}{4}=\\\\=\dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt2}{4}\cdot(1+\sqrt3);\\\\cos\dfrac{7\pi}{12}=cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=cos\dfrac{\pi}{3}\cdot cos\dfrac{\pi}{4}-sin\dfrac{\pi}{3}\cdot sin\dfrac{\pi}{4}=\\\\=\dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt2}{4}\cdot(1-\sqrt3);\\\\cos^4\dfrac{\pi}{12}+cos^4\dfrac{7\pi}{12}=\left[\dfrac{\sqrt2}{4}\cdot(1+\sqrt3)\right]^4+\left[\dfrac{\sqrt2}{4}\cdot(1-\sqrt3)\right]^4=\\\\\dfrac{1}{4^3}\cdot[(1+\sqrt3)^4+(1-\sqrt3)^4]=\dfrac{1}{4^3}\cdot[1+4\cdot\sqrt3+6\cdot3+4\cdot(\sqrt3)^3+(\sqrt3)^4+\\\\+1-4\cdot\sqrt3+6\cdot3-4\cdot(\sqrt3)^3+(\sqrt3)^4]=\dfrac{2}{4^3}\cdot(1+18+9)=\dfrac{7}{8}.$

Green eyes.