Finding absolute extrema of arctangent on a trigonometric interval

Key concept: This is a maximum-minimum problem on a closed interval where a trig identity reduces the expression to a one-variable bound.

Step by step

Step 1: Simplify the trig expression first

We need the absolute maximum and minimum of

$f(x)=\tan^{-1}(\sin x-\cos x), \qquad x\in[0,\pi].$

A powerful observation is that

$\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right).$

So the function becomes

$f(x)=\tan^{-1}\left(\sqrt2\sin\left(x-\frac\pi4\right)\right).$

Since $\tan^{-1}(t)$ is strictly increasing for all real $t$, the extrema of $f(x)$ occur exactly where the inside expression $\sin x-\cos x$ attains its extrema on $[0,\pi]$.

Step 2: Find the maximum and minimum of the inside function

Now study

$g(x)=\sin x-\cos x.$

Using the identity above, or by differentiation,

$g'(x)=\cos x+\sin x.$

Critical points satisfy

$\cos x+\sin x=0 \Rightarrow \tan x=-1.$

On $[0,\pi]$, this gives

$x=\frac{3\pi}{4}.$

Evaluate the values at the endpoints and critical point:

$g(0)= -1,$ $g\left(\frac{3\pi}{4}\right)=\frac{\sqrt2}{2}-\left(-\frac{\sqrt2}{2}\right)=\sqrt2,$ $g(\pi)=0-(-1)=1.$

So the minimum of $g$ is $-1$ and the maximum is $\sqrt2$.

Because $\tan^{-1}$ is increasing, the minimum of $f$ is

$\tan^{-1}(-1)=-\frac\pi4,$

and the maximum of $f$ is

$\tan^{-1}(\sqrt2).$

Step 3: Match the values to the options

The absolute minimum value is

$-\frac\pi4,$

and the absolute maximum value is

$\tan^{-1}(\sqrt2).$

Therefore the answer is the option expressing the minimum and maximum through these exact values.

Key property used

The monotonicity of $\tan^{-1}x$ is the key. Once you know that the outer function is increasing, you do not need to differentiate the composite expression. It is enough to find the range of the inner trig expression on the interval and then pass those endpoint values through arctangent.

Pitfall alert

A frequent mistake is differentiating the whole composition first and then getting lost in a messy expression. This problem is easier if you first simplify $\sin x-\cos x$ into a single sine function. Another common error is to think the maximum or minimum of the arctangent occurs where the inner function has a derivative zero only, ignoring endpoint values. On a closed interval, endpoints matter just as much as critical points. Also, remember that $\tan^{-1}$ preserves order, so the larger inner value gives the larger outer value.

Try different conditions

If the function were $f(x)=\tan^{-1}(\sin x+\cos x)$ on $[0,\pi]$, the same method would work. You would rewrite the inside as $\sqrt2\sin\left(x+\frac\pi4\right)$ or $\sqrt2\cos\left(x-\frac\pi4\right)$ and then find its extrema on the interval. If the interval changed to $[\pi,2\pi]$, the endpoint values would change, and the absolute maximum might shift to a different critical point. The method stays the same, but the actual extreme values depend on the interval.

Further reading

monotonic inverse tangent, trigonometric identity simplification, absolute extrema on interval