# Question ed504

Nov 23, 2016

We start by squaring both sides to get rid of the square root.

(costheta- sin theta)^2 = (sqrt(cos(pi/4 + theta))^2#

${\cos}^{2} \theta - 2 \cos \theta \sin \theta + {\sin}^{2} \theta = \cos \left(\frac{\pi}{4} + \theta\right)$

We now use the identities ${\cos}^{2} \beta + {\sin}^{2} \beta = 1$ and $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ to simplify.

$1 - 2 \cos \theta \sin \theta = \cos \left(\frac{\pi}{4}\right) \cos \theta - \sin \left(\frac{\pi}{4}\right) \sin \theta$

$1 - 2 \cos \theta \sin \theta = \cos \frac{\theta}{\sqrt{2}} - \sin \frac{\theta}{\sqrt{2}}$

As you can see, the two sides aren't equal, so the identity is false.

Hopefully this helps!

Nov 23, 2016

I think the question should be-
Prove that

$\cos \theta - \sin \theta = \sqrt{2} \cos \left(\frac{\pi}{4} + \theta\right)$

LHS

$= \cos \theta - \sin \theta$

$= \sqrt{2} \left(\frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta\right)$

$= \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) \cos \theta - \sin \left(\frac{\pi}{4}\right) \sin \theta\right)$

$= \sqrt{2} \cos \left(\frac{\pi}{4} + \theta\right) = R H S$

Proved