Simplifying and Solving with Inverse Trigonometric Functions: Quick Revision

A brief note on Simplifying and Solving with Inverse Trigonometric Functions. Read it before a test. Try each question at the end before you open its answer.

Substitutions

• For $1 + x^2$: put $x = \tan t$ with $t$ in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
• For $\sqrt{1 - x^2}$: put $x = \sin t$ with $t$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, or $x = \cos t$ with $t$ in $[0, \pi]$.
• The last step: check that the angle is in the branch. If not, use $\pi - u$ or $\pm\pi$.

Standard results

• Half angle: $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x} = \dfrac{1}{2}\tan^{-1}x$ for $x \ne 0$.
• Cosine form: $\cos^{-1}\dfrac{1 - x^2}{1 + x^2} = 2\left|\tan^{-1}x\right|$.
• Sine fold: $\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1}x$ for $|x| \le \dfrac{1}{\sqrt{2}}$, then $\pi - 2\sin^{-1}x$ above and $-\pi - 2\sin^{-1}x$ below.
• Triple angle: $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ for $|x| \le \dfrac{1}{2}$.
• Phase: $\tan^{-1}\dfrac{\cos x - \sin x}{\cos x + \sin x} = \dfrac{\pi}{4} - x$ for $-\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$.

Solving and the domain

• Steps: find the domain, take a sine, cosine or tangent of both sides, solve, and check every root in the original equation.
• One inverse function: $\sin^{-1}u = a$ has a solution only if $a$ is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
• Partner trick: replace $\cos^{-1}x$ by $\dfrac{\pi}{2} - \sin^{-1}x$.
• Domain: the overlap of the conditions of all the inverse functions. It can be one point.
• Range: if all parts rise, or all fall, look at the two ends of the domain.

A worked example

Solve $\tan^{-1}x + \tan^{-1}4x = \dfrac{\pi}{4}$.

1. Take the tangent. $\dfrac{5x}{1 - 4x^2} = 1$, so $4x^2 + 5x - 1 = 0$ and $x = \dfrac{-5 \pm \sqrt{41}}{8}$. The roots are about $0.175$ and $-1.425$.
2. Check. For $x \approx 0.175$ the product $4x^2 \approx 0.12 < 1$, so the sum is $\dfrac{\pi}{4}$. For $x \approx -1.425$ the product is more than $1$ and $x < 0$, so the sum is $-\dfrac{3\pi}{4}$. That root is false. The solution is $x = \dfrac{-5 + \sqrt{41}}{8}$.

Practice with new numbers

A program to try

The program checks the triple angle form at two numbers, the half angle form, and the two roots of $\tan^{-1}x + \tan^{-1}2x = \dfrac{\pi}{4}$.

Mistakes to avoid

• No check: a root of the tangent equation can be false.
• Forgetting the interval: $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ only for $|x| \le \dfrac{1}{2}$.
• A negative inverse cosine: $\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ is $2|\tan^{-1}x|$.
• One domain only: all the inverse functions must be defined together.

Check yourself

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

2. Find the domain of $\cos^{-1}(2x - 3)$.

3. Solve $2\sin^{-1}x = \cos^{-1}x$.

4. Find $\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ at $x = -1$.

5. Is $x = -1$ a solution of $\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x$?

Sources

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