<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] If $a\sin\theta+b\cos\theta=c$
Squaring we get $$a^2\sin^2\theta+b^2\cos^2\theta+2ab\cos\theta\sin\theta=c^2$$
$$\iff a^2(1-\cos^2\theta)+b^2(1-\sin^2\theta)+2ab\cos\theta\sin\theta=c^2$$
$$\iff c^2-a^2-b^2=(a\cos\theta-b\sin\theta)^2$$
Here $\displaystyle a=4,b=3,c=5\implies c^2-a^2-b^2=?$

Alternatively, 
Let $a=r\cos\phi, b=r\sin\phi\  \ \ \ (1)$ where $r\ge0$
So, $\displaystyle a\sin\theta+b\cos\theta=c\implies r\sin(\theta+\phi)=c$
$\displaystyle a\cos\theta-b\sin\theta=r\cos(\theta+\phi)=r\cdot\pm\sqrt{1-\frac{c^2}{r^2}}=\pm\sqrt{r^2-c^2}$
Now, squaring & adding $(1)$  we get $\displaystyle r^2=a^2+b^2$
</s>