Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$ 
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$

I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here

Find the extrema of
$$a\cos(x)+b\sin(x).$$
By canceling the derivative,
$$-a\sin(z)+b\cos(z)=0$$ or$$\tan(z)=\frac ba.$$
Then, with
$$\cos(z)=\pm\frac1{\sqrt{\tan^2(z)+1}}=\pm\frac a{\sqrt{a^2+b^2}},\\\sin(z)=\pm\frac{\tan(z)}{\sqrt{\tan^2(z)+1}}=\pm\frac b{\sqrt{a^2+b^2}},$$
you obtain
$$a\cos(z)+b\sin(z)=\pm\sqrt{a^2+b^2}.$$
Then for all $x$,
$$-\sqrt{a^2+b^2}\le a\cos(x)+b\sin(x)\le\sqrt{a^2+b^2}.$$
