Question

помогите СРОЧНО НУЖНО)A) sin75 и cos75, заменяя 75 на 45 + 30 градусов 
Б) sin15 и cos15, заменяя 15 на 45 - 30 градусов.

Answer 1

А)
$sin75^0=sin(45^0+30^0)=sin45^0cos30^0+sin30^0cos45^0=\ = frac{sqrt{2}}{2}cdotfrac{sqrt{3}}{2}+frac{sqrt{2}}{2}cdotfrac{1}{2}=frac{sqrt{2}}{4}(sqrt{3}+1);\ cos75^0=cos(45^0+30^0)=cos45^0cos30^0-sin30^0sin45^0=\ = frac{sqrt{2}}{2}cdotfrac{sqrt{3}}{2}-frac{sqrt{2}}{2}cdotfrac{1}{2}=frac{sqrt{2}}{4}(sqrt{3}-1);$
Б)
$sin15^0=sin(45^0-30^0)=sin45^0cos30^0-sin30^0cos45^0=\ = frac{sqrt{2}}{2}cdotfrac{sqrt{3}}{2}-frac{sqrt{2}}{2}cdotfrac{1}{2}=frac{sqrt{2}}{4}(sqrt{3}-1);\ cos15^0=cos(45^0-30^0)=cos45^0cos30^0+sin30^0sin45^0=\ = frac{sqrt{2}}{2}cdotfrac{sqrt{3}}{2}+frac{sqrt{2}}{2}cdotfrac{1}{2}=frac{sqrt{2}}{4}(sqrt{3}+1);$