Question

98 б.

Помогите пожалуйста)

Answer 1

$left|begin{array}{ccc}sin^2a&cos^2a&cos2a\sin^2beta &cos^2beta &cos2beta \sin^2gamma &cos^2gamma &cos2gammaend{array}right|=sin^2a(cos^2beta cdot cos2gamma -cos^2gammacdot cos2beta )-\\\-cos^2a(sin^2beta cdot cos2gamma -sin^2gamma cdot cos2beta )+cos2a(sin^2beta cdot cos^2gamma -sin^2gamma cdot cos^2beta )=\\=sin^2aBig (cos^2beta, (cos^2gamma-sin^2gamma)-cos^2gamma(cos^2beta -sin^2beta )Big )-\\-cos^2aBig (sin^2beta(cos^2gamma -sin^2gamma)-sin^2gamma(cos^2beta-sin^2beta)Big )+$

$+Big (cos^2a-sin^2aBig )Big (sin^2beta cdot cos^2gamma -sin^2gamma cdot cos^2beta Big )=\\=sin^2acos^2beta cos^2gamma -sin^2acos^2beta sin^2gamma -sin^2acos^2beta cos^2gamma +sin^2asin^2beta cos^2gamma -\\-cos^2asin^2beta cos^2gamma +cos^2asin^2beta sin^2gamma +cos^2acos^2beta sin^2gamma -cos^2asin^2beta sin^2gamma+\\+cos^2asin^2beta cos^2gamma -cos^2acos^2beta sin^2gamma -sin^2asin^2beta cos^2gamma +sin^2acos^2beta sin^2gamma =\\=0$