<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] Furthermore, another approach.
We can show that $4 \sin \theta + 3 \cos \theta$ has its maximum at $y = 5$. Or, more generally, $a \sin \theta + b \cos \theta$ has a maximum at $y = \sqrt{a^2 + b^2}$. 
We can manipulate the expression $a \sin \theta + b \cos \theta$ to become $\sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin \theta + \frac{b}{\sqrt{a^2 + b^2}}  \cos \theta\right)$
Notice that $\frac{a}{\sqrt{a^2 + b^2}}$ and $\frac{b}{\sqrt{a^2 + b^2}}$ are equivalent to $\cos \alpha$ and $\sin \alpha$ for $\alpha = \tan^{-1} \frac ba$.So now we have $\sqrt{a^2 + b^2} \left(\cos \alpha \sin \theta + \sin \alpha \cos \theta \right) = \sqrt{a^2+b^2} \sin(\alpha + \theta)$, which is maximized when $\alpha + \theta = \pi/2 \rightarrow \theta = \pi / 2 - \alpha$
Thus, we know that $4 \sin \theta + 3 \cos \theta$ is maximized since it equals $5$. Continuing to work generally, we should substitute $\theta = \pi/2 - \alpha$ into $a \cos \theta - b \sin \theta: a \cos(\pi/2 - \alpha) - b \sin (\pi/2 - \alpha) \rightarrow a \sin \alpha - b \cos \alpha$. Recall that $\alpha = \tan^{-1} \frac{b}{a}$, so it follows that $\alpha = \sin^{-1} \frac{b}{\sqrt{a^2 + b^2}} = \cos^{-1} \frac{a}{\sqrt{a^2 + b^2}}$. So finally we have $a \sin \alpha - b \cos \alpha = \frac{ab}{\sqrt{a^2 + b^2}} - \frac{ba}{\sqrt{a^2+b^2}} = \boxed{0}$
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