Question

Преобразовать тригонометрические выражения:
1) 7cos^2 a - 5 + 7sin^2 a = 0
2) (cos x + sin x)^2 - sin2 x
3) (1 + cos2 a)/(2cos a)

Answer 1

$1)\7cos^2alpha-5+7sin^2alpha=7(cos^2alpha+sin^2alpha)-5=7cdot1-5=7-5=2\\2)\(cosx+sinx)^2-sin2x=cos^2x+2sinxcosx+sin^2x-2sinxcosx=1\\3)\frac{1+cos2alpha}{2cosalpha}=frac{1+cos^2alpha-sin^2alpha}{2cosalpha}=frac{1-sin^2alpha+cos^2alpha}{2cosalpha}=frac{cos^2alpha+cos^2alpha}{2cosalpha}\\=frac{2cos^2alpha}{2cosalpha}=cosalpha\\====================================\sin^2x+cos^2x=1to cos^2x=1-sin^2x\\sin2x=2sinxcosx\\cos2x=cos^2x-sin^2x$

Answer 2

:)

Answer 3

$1) 7cos^{2}a-5+7sin^{2}a=(7cos^{2}a+7sin^{2}a)-5= \ 7(cos^{2}a+sin^{2}a)-5=7-5=2 \ 2)(cosx+sinx)^{2}-sin2x=cos^{2}x+sin^{2}x+2sinx*cosx \ -2sinx*cosx=1 \ 3) frac{1+cos2a}{2cosa}=frac{sin^{2}a+cos^{2}+cos^{2}a-sin^{2}a}{2cosa}= frac{2cos^{2}a}{2cosa}=cosa$