Simplifying with a substitution

What you will be able to do

• choose a substitution that removes a root or a fraction inside an inverse trigonometric function;
• simplify the expression to a multiple of an inverse function, and say for which $x$ the result is true;
• correct the result when the angle falls outside the branch.

Why substitute

Look at $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x}$. It looks hard. But $\sqrt{1 + x^2}$ is what the triangle method gives for a tangent, and the triangle method turns it into a secant. Put $x = \tan t$, and the root disappears.

The shape of the expression tells you which substitution to use.

• $\sqrt{1 + x^2}$, or $1 + x^2$: put $x = \tan t$, with $t$ in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
• $\sqrt{1 - x^2}$: put $x = \sin t$ with $t$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, or $x = \cos t$ with $t$ in $[0, \pi]$.
• $\dfrac{1 - x^2}{1 + x^2}$, $\dfrac{2x}{1 + x^2}$, $\dfrac{2x}{1 - x^2}$: put $x = \tan t$.

Always choose $t$ in the branch of the inverse function that $x$ came from. Then $t = \tan^{-1}x$ or $t = \sin^{-1}x$, and the answer can be written back in $x$.

The four steps

1. Substitute and write the expression as a trigonometric function of an angle.
2. Simplify with identities until it is one trigonometric function of one angle, such as $\tan\dfrac{t}{2}$ or $\sin 2t$.
3. Take the inverse. $\tan^{-1}(\tan u) = u$ only if $u$ is in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
4. Check the branch. If $u$ is outside, bring it in with $\pi - u$ or $\pm\pi$, as in the first course.

Two standard results

The half angle form. For $x \ne 0$, put $x = \tan t$ with $t$ in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Then $\sqrt{1 + x^2} = \sec t$, because $\sec t > 0$ in this interval. So

Now $\dfrac{t}{2}$ is in $\left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$, which is inside the branch. So $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x} = \dfrac{t}{2}$, that is

The fold. Put $x = \sin t$ with $t$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Then $\sqrt{1 - x^2} = \cos t$, and $2x\sqrt{1 - x^2} = \sin 2t$. So the expression is $\sin^{-1}(\sin 2t)$ with $2t$ in $[-\pi, \pi]$. Only when $2t$ is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ is the answer $2t$. Otherwise the rule of the first course gives the angle $\pi - 2t$ or $-\pi - 2t$:

The result has three regions.

• For $|x| \le \dfrac{1}{\sqrt{2}}$ it is $2\sin^{-1}x$.
• For $\dfrac{1}{\sqrt{2}} \le x \le 1$ it is $\pi - 2\sin^{-1}x$.
• For $-1 \le x \le -\dfrac{1}{\sqrt{2}}$ it is $-\pi - 2\sin^{-1}x$.

Two more results come from the same method. $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ for $|x| \le \dfrac{1}{2}$, because $\sin 3t = 3\sin t - 4\sin^3 t$. And $\cos^{-1}(4x^3 - 3x) = 3\cos^{-1}x$ for $\dfrac{1}{2} \le x \le 1$, because $\cos 3t = 4\cos^3 t - 3\cos t$ and $3t$ must stay in $[0, \pi]$.

A worked example

Simplify (a) $\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ for every real $x$, and (b) $\tan^{-1}\dfrac{\cos x - \sin x}{\cos x + \sin x}$.

1. (a) Put $x = \tan t$. Then $\dfrac{1 - x^2}{1 + x^2} = \cos 2t$, and the expression is $\cos^{-1}(\cos 2t)$ with $2t$ in $(-\pi, \pi)$. If $x \ge 0$ then $2t$ is in $[0, \pi)$, and the answer is $2t = 2\tan^{-1}x$. If $x < 0$ then $2t$ is negative, and $\cos^{-1}(\cos 2t) = -2t = -2\tan^{-1}x$. In short, the expression equals $2\left|\tan^{-1}x\right|$.
2. (b) Divide the top and the bottom by $\cos x$ (at $x = \dfrac{\pi}{2}$ the value is $-\dfrac{\pi}{4}$ directly, which agrees): $\dfrac{1 - \tan x}{1 + \tan x} = \tan\left(\dfrac{\pi}{4} - x\right)$. The inverse tangent returns $\dfrac{\pi}{4} - x$ when this angle is in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, that is for $-\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$.

Practise with new numbers

Try it in Python

The program compares the fold expression with its three candidate answers, and then the half angle form with half of $\tan^{-1}x$.

Common mistakes

• Using $x = \tan t$ and forgetting the interval of $t$. The branch of $t$ is what makes the answer true.
• Writing $\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1}x$ for all $x$. It holds only for $|x| \le \dfrac{1}{\sqrt{2}}$.
• Dropping the absolute value. $\cos^{-1}\dfrac{1 - x^2}{1 + x^2} = 2\left|\tan^{-1}x\right|$, which is never negative.
• A wrong half angle. $\dfrac{1 - \cos t}{\sin t} = \tan\dfrac{t}{2}$, so the result has $\dfrac{t}{2}$, not $2t$.

Remember these

• Use $x = \tan t$ for $1 + x^2$, and $x = \sin t$ or $x = \cos t$ for $\sqrt{1 - x^2}$.
• $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x} = \dfrac{1}{2}\tan^{-1}x$ for $x \ne 0$.
• $\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1}x$ only for $|x| \le \dfrac{1}{\sqrt{2}}$. Beyond that the answer is folded by $\pi$.
• $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ only for $|x| \le \dfrac{1}{2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which substitution removes the root in $\sqrt{1 + x^2}$?

2. Simplify $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x}$.

3. Find $\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ at $x = -1$, and compare it with $2\tan^{-1}x$.

4. Write $\sin^{-1}\dfrac{24}{25}$ in terms of $\sin^{-1}\dfrac{4}{5}$ and $\pi$.

5. For which $x$ is $\tan^{-1}\dfrac{\cos x - \sin x}{\cos x + \sin x} = \dfrac{\pi}{4} - x$?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.