Question

cos theta + sin theta =√2 cos theta show that cos theta - sin theta =√2 sin theta

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,

Given:

$\cos(theta) + \sin(theta) = \sqrt{2} \cos(theta)$

Let theta be any variable, here, theta = b. Hence it will be like this:

$= \cos(b) + \sin(b) = \sqrt{2} \cos(b)$

Squaring on left hand side and right hand sides, we get,,,,

$= {( \cos(b) + \sin(b) ) }^{2} = {( \sqrt{2} \cos(b) ) }^{2}$
$= ( \cos(b) \times \cos(b) ) + ( \sin(b) \times \sin(b) ) + ( \cos(b) \times \sin(b) ) + ( \cos(b) \times \sin(b) )$

Hence,, we get,,

$= \cos(2b) + \sin(2b) + 2 \cos(b) \sin(b) = 2 \cos(2b)$
$= 2 \sin(b) \cos(b) = 2 \cos(2b) - \cos(2b) - \sin(2b) \\ \\ = 2 \sin(b) \cos(b) = \cos(2b) - \sin(2b)$
$= 2 \sin(b) \cos(b) = ( \cos(b) + \sin(b) ) \times ( \cos(b) - \sin(b) ) \\ \\ \\ = 2 \sin(b) \cos(b) = ( \sqrt{2} \cos(b)) \times ( \cos(b) - \sin(b) )$

Note: Since,, the equation was cos (b) + sin (b) = root2 cos (b),, the above value was substituted.

Concluding further and shifting the nominator into denominator,,, the 2 and root2 get cancelled and cos b gets cancelled out, let me show it for better understanding

$= \frac{2 \sin(b) \cos(b) }{ \sqrt{2} \cos(b) } = \cos(b) - \sin(b) \\ \\ \\ = \frac{ \sqrt{2} \times \sqrt{2} \sin(b) \cos(b) }{ \sqrt{2} \cos(b) } = \cos(b) - \sin(b)$

Therefore,,,,,, we get,,,,,

$\sqrt{2} \sin(b) = \cos(b) - \sin(b)$

Since,,, b = theta or theta = b

$\cos(theta) - \sin(theta) = \sqrt{2} \sin(theta)$

Hence shown or proved according to given trigonometric identities.

HOPE IT HELPS AND SOLVES YOUR DOUBTS RELATING TRIGONOMETRY AND IT'S IDENTITIES!!!!