In the course of evaluating the antiderivative

,

I have stumbled across a very curious trigonometric identity:

if ,

if ,

if ,

where and are the unique values so that

,

.

I will now show that and . Indeed, it’s apparent that these have to be the two transition points because these are the points where is undefined. However, it would be more convincing to show this directly.

To show that , I need to show that

.

I could do this with a calculator…

…but that would be cheating.

Instead, let and , so that

,

.

Indeed, by SOHCAHTOA, the angles and can be represented in the figure below:

The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly . I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing , the missing side is

Next, for the small right triangle containing , the missing side is

So let me redraw the figure, eliminating the altitude from the previous figure:

Notice that the condition of the Pythagorean theorem is satisfied, since

,

or

.

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so . In other words, , as required.

To show that , I will show that the function is an odd function using the fact that is also an odd function:

.

Therefore, , and so .

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