<s>[INST] Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ  = 5$$
Find 
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
 [/INST] Its simple:
Given
$$4\sin\theta + 3\cos\theta=5$$
Squaring on both sides gives
$$16\sin^2\theta +24\sin\theta \cos\theta+9\cos^2\theta=25$$
$$16-16\cos^2\theta+24\sin\theta \cos\theta+9-9\sin^2\theta=25$$
$$16\cos^2\theta-24\sin\theta \cos\theta+9\sin^2\theta=0$$
$$(4\cos\theta-3\sin\theta)^2=0$$
$$4\cos\theta-3\sin\theta=0$$
Hope it helps!.
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