Question

If cosθ + sin θ = √2 cosθ, prove that cosθ - sinθ =√2 sinθ.

Answer 1

$\textbf{Given:}$

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

$\text{Squaring on both sides, we get}$

$(cos\theta+sin\theta)^2=2\,cos^2\theta$

$\text{using}$

$\boxed{\bf(a+b)^2=a^2+b^2+2ab}$

$cos^2\theta+sin^2\theta+2\,sin\theta\,cos\theta=2(1-\,sin^2\theta)$

$1+2\,sin\theta\,cos\theta=2-2\,sin^2\theta$

$\text{Rearranging terms we get}$

$2\,sin^2\theta+2\,sin\theta\,cos\theta=2-1$

$2\,sin^2\theta=1-2\,sin\theta\,cos\theta$

$2\,sin^2\theta=cos^2\theta+sin^2\theta-2\,sin\theta\,cos\theta$

$\text{using}$

$\boxed{\bf(a-b)^2=a^2+b^2-2ab}$

$2\,sin^2\theta=(cos\theta-sin\theta)^2$

$\text{Taking square root on both sides, we get}$

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

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