Question

98 б.

Помогите пожалуйста)

Answer 1

$\left|\begin{array}{ccc}sin^2a&cos^2a&cos2a\\sin^2\beta &cos^2\beta &cos2\beta \\sin^2\gamma &cos^2\gamma &cos2\gamma\end{array}\right|=sin^2a(cos^2\beta \cdot cos2\gamma -cos^2\gamma\cdot cos2\beta )-\\\\\\-cos^2a(sin^2\beta \cdot cos2\gamma -sin^2\gamma \cdot cos2\beta )+cos2a(sin^2\beta \cdot cos^2\gamma -sin^2\gamma \cdot cos^2\beta )=\\\\=sin^2a\Big (cos^2\beta\, (cos^2\gamma-sin^2\gamma)-cos^2\gamma(cos^2\beta -sin^2\beta )\Big )-\\\\-cos^2a\Big (sin^2\beta(cos^2\gamma -sin^2\gamma)-sin^2\gamma(cos^2\beta-sin^2\beta)\Big )+$

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