Equations and applications

The identities of the previous subtopics turn inverse trigonometric equations into ordinary algebraic ones. The procedure is universal: combine all inverse trig terms into one using identities, take the appropriate trigonometric function of both sides, solve for the variable, and finally check that the solutions lie in the original domains. Skipping the last step is the most common source of wrong answers.

A standard strategy

1. Move all inverse trig terms to one side.
2. Apply addition/multi-angle identities until only one inverse trig term remains.
3. Take the corresponding trigonometric function of both sides.
4. Solve algebraically.
5. Verify each candidate solution against the original domain constraints.

Examples requiring branch awareness

A candidate $x$ produced by the algebra may not satisfy the principal range requirements. For instance, if you solve $\sin^{-1} x + \sin^{-1}(2x) = \pi/2$ by taking $\sin$, you may produce extraneous roots.

Worked examples

Example 1. Solve $\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4$.

Apply the addition formula (assuming $6x^2 < 1$, i.e., $|x| < 1/\sqrt 6$):

$\tan^{-1}\tfrac{5x}{1 - 6x^2} = \pi/4$, so $\tfrac{5x}{1 - 6x^2} = 1$. This gives $6x^2 + 5x - 1 = 0$, factoring as $(6x - 1)(x + 1) = 0$. Candidates $x = 1/6$ or $x = -1$. Check $|x| < 1/\sqrt 6 \approx 0.408$: $x = 1/6 \approx 0.167$ passes; $x = -1$ fails (and indeed $\tan^{-1}(-2) + \tan^{-1}(-3)$ is negative, not $\pi/4$).

Final answer: $x = 1/6$.

Example 2. Solve $\sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x$.

Let $\theta = \sin^{-1} x$ so $x = \sin \theta$. Use $\cos^{-1} x = \pi/2 - \sin^{-1} x = \pi/2 - \theta$. The equation becomes $\theta + \sin^{-1}(1 - x) = \pi/2 - \theta$, i.e., $\sin^{-1}(1 - x) = \pi/2 - 2\theta$, so $1 - x = \sin(\pi/2 - 2\theta) = \cos 2\theta = 1 - 2 \sin^2 \theta = 1 - 2x^2$. Hence $x = 2x^2$, giving $x(2x - 1) = 0$: $x = 0$ or $x = 1/2$.

Verify both: $x = 0$: $\sin^{-1} 0 + \sin^{-1} 1 = 0 + \pi/2 = \pi/2 = \cos^{-1} 0$. $\checkmark$. $x = 1/2$: $\sin^{-1}(1/2) + \sin^{-1}(1/2) = \pi/6 + \pi/6 = \pi/3 = \cos^{-1}(1/2)$. $\checkmark$.

Both work.

Example 3. Solve $\tan^{-1}\tfrac{1 - x}{1 + x} = \tfrac{1}{2} \tan^{-1} x$ for $x > 0$.

Use the identity $\tan^{-1}\tfrac{1 - x}{1 + x} = \tan^{-1} 1 - \tan^{-1} x = \pi/4 - \tan^{-1} x$ (valid for $x > -1$).

Equation: $\pi/4 - \tan^{-1} x = \tfrac{1}{2} \tan^{-1} x$, so $\pi/4 = \tfrac{3}{2} \tan^{-1} x$, hence $\tan^{-1} x = \pi/6$, so $x = 1/\sqrt 3$.

Example 4. Solve $\cos^{-1} x + \cos^{-1}(2x) = \pi/3$.

Take cosine. Let $\alpha = \cos^{-1} x, \beta = \cos^{-1}(2x)$. Then $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = 2x^2 - \sqrt{1 - x^2}\sqrt{1 - 4x^2} = 1/2$.

Rearrange: $\sqrt{(1 - x^2)(1 - 4x^2)} = 2x^2 - 1/2$. Square: $(1 - x^2)(1 - 4x^2) = (2x^2 - 1/2)^2 = 4x^4 - 2x^2 + 1/4$.

Left: $1 - 5x^2 + 4x^4$. Equate: $1 - 5x^2 + 4x^4 = 4x^4 - 2x^2 + 1/4$, so $1 - 5x^2 = -2x^2 + 1/4$, giving $3x^2 = 3/4$, $x^2 = 1/4$, $x = \pm 1/2$.

Check the squaring step required $2x^2 - 1/2 \ge 0$, i.e., $|x| \ge 1/2$. Both $\pm 1/2$ are on the boundary. Verify: $x = 1/2$: $\cos^{-1}(1/2) + \cos^{-1}(1) = \pi/3 + 0 = \pi/3$. $\checkmark$. $x = -1/2$: $\cos^{-1}(-1/2) + \cos^{-1}(-1) = 2\pi/3 + \pi = 5\pi/3 \neq \pi/3$. So only $x = 1/2$ works.

Example 5. Find the number of real solutions of $\tan^{-1} x + \tan^{-1}(1 - x) = \tan^{-1}\tfrac{7}{9}$.

LHS: provided $x(1 - x) < 1$, this equals $\tan^{-1}\tfrac{x + 1 - x}{1 - x(1 - x)} = \tan^{-1}\tfrac{1}{1 - x + x^2}$. Setting $\tfrac{1}{1 - x + x^2} = \tfrac{7}{9}$ gives $9 = 7(1 - x + x^2)$, so $7x^2 - 7x - 2 = 0$. Discriminant $49 + 56 = 105 > 0$, two roots.

Check the branch condition $x(1 - x) < 1$ , always true since $x(1 - x) \le 1/4 < 1$. So both roots satisfy.

Two solutions.

Example 6. A ladder of length $13$ m leans against a wall, with foot $5$ m from the base. The angle the ladder makes with the ground is $\theta = \cos^{-1}(5/13)$. Express $\theta$ as $\sin^{-1}$ and as $\tan^{-1}$.

Using a $5$-$12$-$13$ triangle: $\sin \theta = 12/13$, $\tan \theta = 12/5$. So $\theta = \sin^{-1}(12/13) = \tan^{-1}(12/5) = \cos^{-1}(5/13)$.

Try it yourself

1. Solve $\tan^{-1} x + 2 \cot^{-1} x = 2\pi/3$.
2. Solve $\sin^{-1}(x) + \sin^{-1}(2x) = \pi/3$.
3. Solve $\cos^{-1}(x) + \cos^{-1}(2x) = \pi/2$.
4. Solve $\tan^{-1}(x - 1) + \tan^{-1}(x + 1) = \pi/4$.
5. Solve $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$.
6. Show that $\tan^{-1}\tfrac{1}{3} + \tan^{-1}\tfrac{1}{5} + \tan^{-1}\tfrac{1}{7} + \tan^{-1}\tfrac{1}{8} = \pi/4$.
7. If $\sin^{-1} x + \sin^{-1} y = \pi/3$ and $\cos^{-1} x - \cos^{-1} y = \pi/6$, find $x$ and $y$.
8. From the top of a tower of height $h$, the angle of depression to a point at distance $d$ is $\theta = \tan^{-1}(h/d)$. If $h = 30, d = 40$, find $\theta$.
9. Solve $\sin^{-1}(\tfrac{5}{x}) + \sin^{-1}(\tfrac{12}{x}) = \pi/2$ for $x$.
10. Show that $\tan^{-1}\tfrac{x + a}{1 - ax} = \tan^{-1} x + \tan^{-1} a$, with appropriate conditions.
11. A boy walks $50$ m east and then $50\sqrt 3$ m north. Express the angle from his start direction to his final position using $\tan^{-1}$.
12. Solve $\tan^{-1}\tfrac{x - 1}{x - 2} + \tan^{-1}\tfrac{x + 1}{x + 2} = \pi/4$.
13. Find all $x \in \mathbb{R}$ satisfying $\sin^{-1} x = 2 \tan^{-1}\tfrac{x}{1 + \sqrt{1 - x^2}}$.
14. Solve $\cos^{-1}(1 - x) + \cos^{-1}(x) = \pi/2$.

Pitfalls / Tricks

• After any squaring or trig-function step, always re-verify candidates in the original equation. Spurious roots are routine.
• A single forgotten branch correction $\pm \pi$ can turn a correct answer into a wrong one.
• For applied (geometry) problems, draw the triangle first and read off ratios , don't reach for formulas blindly.
• A function like $\tan^{-1} \tfrac{x - 1}{x + 1}$ often equals $\tan^{-1} x - \pi/4$; recognising such patterns shortcuts many problems.

This concludes the chapter. Inverse trigonometric functions return in nearly every chapter to come , calculus, integration, complex analysis. Master the principal branches and the addition formulas first.