Question

What is the maximum value of the function sin x + cos x?

Answer 1

Let f(x) = sin x + cos x

=f'(x)=cos x-sinx

f'(x)=0=sinx=cos x=tanx=1=\(\frac{\pi}{4},\frac{5\pi}{4}\)....,

f''(x)=-sinx-cos x=-(sinx+cos x)

Now, f\(\times\)(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first

quadrant. Then, f\(\times\)(x) will be negative when x∈(0,\(\frac{\pi}{2}\)).

Thus, we consider x=\(\frac{\pi}{4}\).

f\(\times\)(\(\frac{\pi}{4}\))=-(sin \(\frac{\pi}{4}\)+cos \(\frac{\pi}{4}\))=-(\(\frac{2}{\sqrt2}\))=\(-\sqrt2<0\)

∴ By the second derivative test, f will be the maximum at x=π/4. and the maximum value of f is f(\(\frac{\pi}{4}\))=sin \(\frac{\pi}{4}\).+cos \(\frac{\pi}{4}\)=\(\frac{1}{\sqrt2}\times \frac{1}{\sqrt2}\)=\(\frac{2}{\sqrt2}\)=\(\sqrt2\)