#363 (докажите , что они равны) пожалуйста помогите

Question 1

Question 2

Решите задачу:

$sin( \alpha + \beta ) * sin ( \alpha - \beta ) = \\ =(sin \alpha *cos \beta +cos \alpha *sin \beta )*(sin \alpha *cos \beta - cos \alpha* sin \beta ) = \\ =sin^2 \alpha cos^2 \beta +cos \alpha sin \alpha sin \beta cos \beta - sin \alpha cos \alpha cos \beta sin \beta -cos^2 \alpha sin^2 \beta \\ = sin^2 \alpha cos^2 \beta - cos^2 \alpha sin^2 \beta = \\ =sin^2 \alpha *(1-sin^2 \beta ) - sin^2 \beta *(1-sin^2 \alpha ) = \\ = sin^2 \alpha -sin^2 \alpha sin^2 \beta +sin^2 \beta +sin^2 \alpha sin^2 \beta$
$= sin^2 \alpha +sin^2 \beta$

$cos( \alpha + \beta )*cos ( \alpha - \beta ) = \\ =(cos \alpha cos \beta -sin \alpha sin \beta )* (cos \alpha cos \beta +sin \alpha sin \beta ) = \\ =cos^2 \alpha cos^2 \beta -sin \alpha sin \beta cos \alpha cos \beta+cos \alpha sin \alpha cos \beta sin \beta -sin^2 \alpha sin^2 \beta \\ =cos^2 \alpha cos^2 \beta-sin^2 \alpha sin^2 \beta = \\ = cos^2 \alpha *(1-sin^2 \beta )-sin^2 \beta (1-cos^2 \alpha )= \\ =cos^2 \alpha -sin^2 \beta cos^2 \alpha -sin^2 \beta +sin^2 \beta cos^2 \alpha = \\$
$=cos^2 \alpha -sin^2 \beta$

$sin( \alpha + \beta )*cos( \alpha - \beta ) = \\ =(sin \alpha cos \beta +cos \alpha sin \beta )*(cos \alpha cos \beta +sin \alpha sin \beta ) = \\ = sin \alpha cos \alpha cos^2 \beta +sin \beta cos \beta cos^2 \alpha +sin \beta cos \beta sin^2 \alpha +sin \alpha cos \alpha sin^2 \beta \\ =sin \alpha cos \alpha (cos^2 \beta +sin^2 \beta )+sin \beta cos \beta (cos^2 \alpha +sin^2 \alpha )= \\ =sin \alpha cos \alpha *1 + sin \beta cos \beta *1 = \\ =sin \alpha cos \alpha +sin \beta cos \beta$