Question

Алгебра, 10
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Answer 1

Ответ:

12

$\frac{ \cos {}^{2} ( 2\alpha ) - 4 \sin {}^{2} ( \alpha ) \cos {}^{2} ( \alpha ) }{ctg (2\alpha ) - \sin( 4\alpha ) } = \\ = \frac{ \cos {}^{2} ( 2\alpha ) - (2 \sin( \alpha ) \cos( \alpha )) {}^{2} }{ \frac{ \cos(2 \alpha ) }{ \sin( 2\alpha ) } - 2 \sin( 2\alpha ) \cos(2 \alpha ) } = \\ = ( \cos {}^{2} (2 \alpha ) - \sin {}^{2} ( 2\alpha ) ) \times \frac{ \sin(2 \alpha ) }{ \cos( 2\alpha ) - 2 \sin {}^{2} ( 2\alpha ) \cos( 2\alpha ) } = \\ = ( \cos {}^{2} (2 \alpha ) - \sin {}^{2} ( 2\alpha )) \times \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) (1 - 2 \sin {}^{2} (2 \alpha )) } = \\ = \cos( 4\alpha ) \times \frac{ \sin( 2\alpha ) }{ \cos( 2\alpha ) \times \cos( 4\alpha ) } = tg(2 \alpha )$

13

$\frac{ctg \alpha - \sin(2 \alpha ) }{ \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) } = \frac{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - 2 \sin( \alpha ) \cos( \alpha ) }{ \cos( 2\alpha ) } = \\ = \frac{ \cos( \alpha ) - 2 \sin {}^{2} ( \alpha ) \cos( \alpha ) }{ \sin( \alpha ) } \times \frac{1}{ \cos(2 \alpha ) } = \\ = \frac{ \cos( \alpha ) (1 - 2 \sin {}^{2} ( \alpha )) }{ \sin( \alpha ) \cos( 2\alpha ) } = \frac{ \cos( \alpha ) \cos( 2\alpha ) }{ \sin( \alpha ) \cos( 2\alpha ) } = ctg \alpha$