Question

689. Дано: $\cos \alpha = \frac{1}{ \sqrt{2} } \cos \beta = \frac{1}{3}$
а и в - углы четырех четвертей. Вычислить:
1)
2)
3)

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$\sin( \alpha - \beta )$
$\cos( \alpha - \beta )$
$\tan( \alpha - \beta )$

Answer 1

$\cos( \alpha ) = \frac{1}{ \sqrt{2} } \\ \cos( \beta ) = \frac{1}{3} \\ \\ \sin( \alpha ) = \sqrt{1 - { \cos }^{2} \alpha } \\ \sin( \alpha ) = - \sqrt{1 - \frac{1}{2} } = - \frac{1}{ \sqrt{2} } \\ \\ \sin( \beta ) = \sqrt{1 - \cos ^{2} ( \beta ) } \\ \sin( \beta ) = - \sqrt{1 - \frac{1}{9} } = - \frac{ \sqrt{8} }{3} = - \frac{2 \sqrt{2} }{3}$

1

$\sin( \alpha - \beta ) = \\ = \sin( \alpha ) \cos( \beta ) - \sin( \beta ) \cos( \alpha ) = \\ = - \frac{1}{ \sqrt{2} } \times \frac{1}{3} - ( - \frac{2 \sqrt{2} }{3} \times \frac{1}{ \sqrt{2} } ) = \\ = - \frac{1}{3 \sqrt{2} } + \frac{2}{3} = \frac{2 \sqrt{2} - 1 }{3 \sqrt{2} }$

2

$\cos( \alpha - \beta ) = \\ = \cos( \alpha ) \cos( \beta ) + \sin( \alpha ) \sin( \beta ) = \\ = \frac{1}{ \sqrt{2} } \times \frac{1}{3} + ( - \frac{ 1}{ \sqrt{2} } \times \frac{2 \sqrt{2} }{3} ) = \\ = \frac{1}{3 \sqrt{2} } - \frac{2}{3} = \frac{1 - 2 \sqrt{2} }{3 \sqrt{2} }$

$tg( \alpha - \beta ) = \frac{ \sin( \alpha - \beta ) }{ \cos( \alpha - \beta ) } = \\ = \frac{2 \sqrt{2} - 1 }{3 \sqrt{2} } \times \frac{3 \sqrt{2} }{1 - 2 \sqrt{2} } = \\ = - \frac{1 - 2 \sqrt{2} }{1 - 2 \sqrt{2} } = - 1$