# Find maxima and minima of f(x)= 5sinx + 5cosx on a interval of [0,2pi]?

May 13, 2018

There's

• a local maximum at $\left(\frac{\pi}{2} , 5\right)$ and
• a local minimum at $\left(\frac{3 \pi}{2} , - 5\right)$

#### Explanation:

$\textcolor{\mathrm{da} r k b l u e}{\sin \left(\frac{\pi}{4}\right)} = \textcolor{\mathrm{da} r k b l u e}{\cos \left(\frac{\pi}{4}\right)} = \textcolor{\mathrm{da} r k b l u e}{1}$

$f \left(x\right) = 5 \sin x + 5 \cos x$
$\textcolor{w h i t e}{f \left(x\right)} = 5 \left(\textcolor{\mathrm{da} r k b l u e}{1} \cdot \sin x + \textcolor{\mathrm{da} r k b l u e}{1} \cdot \cos x\right)$
$\textcolor{w h i t e}{f \left(x\right)} = 5 \left(\textcolor{\mathrm{da} r k b l u e}{\cos \left(\frac{\pi}{4}\right)} \cdot \sin x + \textcolor{\mathrm{da} r k b l u e}{\sin \left(\frac{\pi}{4}\right)} \cdot \cos x\right)$

Apply the compound angle identity for the sine function

$\sin \left(\alpha + \beta\right) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta$

$\textcolor{b l a c k}{f \left(x\right)} = 5 \cdot \sin \left(\frac{\pi}{4} + x\right)$

Let $x$ be the $x -$coordinate of local extrema of this function.

$5 \cdot \cos \left(\frac{\pi}{4} + x\right) = f ' \left(x\right) = 0$
$\frac{\pi}{4} + x = \frac{\pi}{2} + k \cdot \pi$ where $k$ an integer.
$x = - \frac{\pi}{2} + k \cdot \pi$

$x \in \left\{\frac{\pi}{2} , \frac{3 \pi}{2}\right\}$

• $f \left(\frac{\pi}{2}\right) = 5 \cdot \sin \left(\frac{\pi}{2}\right) = 5$,
hence there's a local maximum at $\left(\frac{\pi}{2} , 5\right)$
• $f \left(\frac{\pi}{2}\right) = 5 \cdot \sin \left(\frac{3 \pi}{2}\right) = - 5$,
hence there's a local minimum at $\left(\frac{\pi}{2} , - 5\right)$