Question

The maximum value of f(x) = sin x + cos x is _______________.

Answer 1

The given function is $f\left(x\right)=\sin x+\cos x$.

$f\left(x\right)=\sin x+\cos x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\cos x-\sin x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \cos x=\sin x$

$\Rightarrow \tan x=1$

$\Rightarrow x=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$ (Let us only consider the values of x ∈ [0, 2$\mathrm{\pi }$])

Now,

$f\text{'}\text{'}\left(x\right)=-\sin x-\cos x$

At $x=\frac{5\mathrm{\pi }}{4}$, we have

$f\text{'}\text{'}\left(\frac{5\mathrm{\pi }}{4}\right)$ $=-\sin \frac{5\mathrm{\pi }}{4}-\cos \frac{5\mathrm{\pi }}{4}$ $=-\left(-\frac{1}{\sqrt{2}}\right)-\left(-\frac{1}{\sqrt{2}}\right)$ $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}}$ $=\sqrt{2}$ > 0

So, $x=\frac{5\mathrm{\pi }}{4}$ is the point of local minimum of f(x).

At $x=\frac{\mathrm{\pi }}{4}$, we have

$f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ $=-\sin \frac{\mathrm{\pi }}{4}-\cos \frac{\mathrm{\pi }}{4}$ $=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$ $=-\frac{2}{\sqrt{2}}$ $=-\sqrt{2}$ < 0

So, $x=\frac{\mathrm{\pi }}{4}$ is the point of local maximum of f(x).

∴ Maximum value of f(x) $=f\left(\frac{\mathrm{\pi }}{4}\right)$ $=\sin \frac{\mathrm{\pi }}{4}+\cos \frac{\mathrm{\pi }}{4}$ $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}}$ $=\sqrt{2}$

Thus, the maximum value of $f\left(x\right)=\sin x+\cos x$ is $\sqrt{2}$.


The maximum value of f(x) = sinx + cosx is $\underline{\sqrt{2}}$.