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 $\frac{\pi}{4}$=$\frac{1}{\sqrt2}\times \frac{1}{\sqrt2}$=$\frac{2}{\sqrt2}$=$\sqrt2$