Trig1.1

If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta\,$ prove that $\cos\theta-\sin\theta=\sqrt{2}\sin\theta\,$

$\sin\theta=\sqrt{2}\cos\theta-\cos\theta=(\sqrt{2}-1)\cos\theta\,$

Multiplying both sides by $\sqrt{2}+1\,$

$(\sqrt{2}+1)\sin\theta=(\sqrt{2}+1)(\sqrt{2}-1)\cos\theta\,$

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

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

Hence the required follows.