Question

Find the maximum value of $\sin x+\cos x$

Answer 1

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

Now, $f^{\prime }\left(x\right)=\cos x-\sin x$

And $f^{\prime \prime }\left(x\right)=-(\sin x+\cos x)$.

Now for maximum or minimum value of $f\left(x\right)$ we must have,

$f^{\prime }\left(x\right)=0$

or, $\cos x-\sin x=0$

or, $\tan x=1$

or, $x=n\pi +\frac{\pi }{4}$. [ Where $n$ is an integer]

Now, for $n=0$ we have, $x=\frac{\pi }{4}$.

Now for $x=\frac{\pi }{4}$, we have $f^{\prime \prime }\left(x\right)<0$.

So we have maximum value for $x=\frac{\pi }{4}$.

So the maximum value of $f\left(x\right)=\sin {45}^o+\cos {45}^o=\sqrt{2}$.