Prove that $-4\leq5\cos\theta+3\cos(\theta+\frac{\pi}{3})+3\leq10$

Question

Prove that $$\color\red{-4}\leq5\cos\theta+3\cos(\theta+\frac{\pi}{3})+3\leq\color\red{10}$$

My attempt:-

I simplified the equation to

$$\begin{align} &\;\;\phantom{=} 5\cos\theta+3\cos(\theta+60^0)+3 \\[4pt] &= 5\cos\theta+3(\cos\theta\cos60^0-\sin\theta\sin60^0)+3\\[4pt] &= \frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta+3 \end{align}$$

I find no way to continue and eliminate the $\sqrt3$ and get the exact values of $-4$ and $10$ although solving the above equation gives me a near approximation.

How do I get the exact value (best without calculus but it may be accepted)?

Thanks for any help!!

Answer 1

You can use this $$f(x)=a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\alpha)$$ where $\cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. So clearly $$-\sqrt{a^2+b^2}\le f(x) \le \sqrt{a^2+b^2}$$ Now in your given equation $$g(x)=\frac{13}{2}\cos(x)-\frac{3\sqrt{3}}{2}\sin(x)+3=h(x)+3$$ where $h(x)=\frac{13}{2}\cos(x)-\frac{3\sqrt{3}}{2}\sin(x)$

Clearly from above argument, $-7\le h(x) \le 7$

So $$-4\le g(x) \le 10$$ which is the answer.

Hope this will be helpful !