Question

if cos + sin = √2cos prove that cos - sin =√2 sin

Answer 1

hope this will help...

Answer 2

Answer:

$cos \theta -sin\theta =\sqrt{2}sin \theta$ is proved

Step-by-step explanation:

Given  $cos\theta + sin\theta =\sqrt{2} cos\theta$

Squaring on both sides,  

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

Expanding the square

$\implies cos^{2} \theta + sin^{2} \theta+2cos\theta sin\theta ={2} cos^{2} \theta$

$\implies (1-2)cos^{2} \theta+ sin^{2} \theta+2cos\theta sin\theta =0$

$\implies - cos^{2} \theta+ sin^{2} \theta+2cos\theta sin\theta =0$

$\implies cos^{2} \theta- sin^{2} \theta+2cos\theta sin\theta =0$

Adding ${2sin^{2} \theta}$ on both sides,

$\implies - cos^{2} \theta+ sin^{2} \theta+2cos\theta sin\theta+{2sin^{2} \theta} ={2sin^{2} \theta}$

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

Taking on root both sides

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

$\implies cos \theta -sin\theta =\sqrt{2}sin \theta$

Hence proved