Question

solve this problem..... ​

Answer 1

$HEYA \: \\ \\ put \: x \: = cos( \alpha ) \\ \alpha = \cos {}^{ - 1} ( \alpha ) \\ \\ \sin {}^{ - 1} ( 2 \times \cos( \alpha ) \sqrt{ \sin {}^{2} ( \alpha ) } \\ becoz \: (1 - \cos {}^{2} ( \alpha ) = \sin {}^{2} ( \alpha ) ) \\ \\ \sin {}^{ - 1} ( ( \sin(2 \alpha ) ) \\ becoz \: \: 2 \times sin( \alpha ) \times \cos( \alpha ) = \sin(2 \alpha ) \\ \\ 2 \alpha \\ \\ 2 \cos {}^{ - 1} ( \alpha )$

Answer 2

welcome to the concept of inverse Trigonometry

formula used:

$\sin {}^{ - 1} ( \sin(x) ) = x$

solution:

we have to proof

$\sin {}^{ - 1} (2x \sqrt{1 - x {}^{2} } ) = 2 \cos {}^{ - 1} (x)$

LHS

$= \sin {}^{ - 1} (2x \sqrt{1 - x {}^{2} } )$

let x= cosy

THUS , cos inverse x = y

$= \sin {}^{ - 1} (2 \cos(y) \sqrt{1 - \cos {}^{2} (y) } )$

$= \sin {}^{ - 1} (2 \cos(y) \sin(y) )$

$= \sin {}^{ - 1} ( \sin(2y) )$

$= 2y$

$= 2 \cos {}^{ - 1} (x)$

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