The maximum and minimum values of the function $f:[R \rightarrow[R$ defined by $f(x)=5 \cos x+3 \cos…
The maximum and minimum values of the function $f:[R \rightarrow[R$ defined by $f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8$ for all $x \in[R$, are respectively.
15, 1
8, - 8
-7, - 15
1, - 15
Solution
$
\begin{aligned}
& f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8 \\
& =5 \cos x+3\left[\cos x \cdot \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}\right]+8 \\
& =5 \cos x+3\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]+8 \\
& =\frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+8
\end{aligned}
$
We know that, $a \cos x-b \sin x$ is lie between $\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]$
$
\begin{gathered}
-\sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{-3 \sqrt{3}}{2}\right)^2}+8 \leq f(x) \\
\leq \sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{-3 \sqrt{3}}{2}\right)^2}+8 \\
-7+8 \leq f(x) \leq 7+8 \\
1 \leq f(x) \leq 15
\end{gathered}
$
So, maximum value of $f(x)$ is 15 and minimum value is 1