25 Solved Examples Across the Chapter

The problems below run in rough order of difficulty: principal values and wrap-around evaluations (Examples 1 to 9), the sum-and-simplification proofs built on the triangle method (10 to 17), and equations plus the harder closing classics (18 to 25).

Two tools do almost all the work in this chapter: the branch check (is the angle inside the principal range?) and the triangle method (draw a right triangle to convert between $\sin^{-1}, \cos^{-1}, \tan^{-1}$ of related ratios). Watch how often each appears below.

Examples 1 to 9 — Principal Values and the Triangle Method

Example 1 — Wrap-around past $2\pi$

Find $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$.

Step 1 — reduce by full turns: $\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$, so $\cos\frac{13\pi}{6} = \cos\frac{\pi}{6}$.

Step 2 — branch check: $\frac{\pi}{6} \in [0, \pi]$ ✓.

Answer: $\dfrac{\pi}{6}$.

Example 2 — Wrap-around with period $\pi$

Find $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$.

Step 1 — subtract the period: $\tan\frac{7\pi}{6} = \tan\left(\frac{7\pi}{6} - \pi\right) = \tan\frac{\pi}{6}$.

Step 2 — branch check: $\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ✓.

Answer: $\dfrac{\pi}{6}$ — not $\frac{7\pi}{6}$.

Example 3 — Negative inputs to the reciprocal functions

Find the principal values of (i) $\mathrm{cosec}^{-1}(-2)$ and (ii) $\sec^{-1}(-\sqrt 2)$.

Step 1 — (i): $\mathrm{cosec}\,y = -2 \iff \sin y = -\frac{1}{2}$; the odd trio gives negatives: $y = -\frac{\pi}{6}$.

Step 2 — (ii): $\sec y = -\sqrt 2 \iff \cos y = -\frac{1}{\sqrt 2}$; the $[0, \pi]$ family goes to the second quadrant: $y = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Answer: (i) $-\dfrac{\pi}{6}$, (ii) $\dfrac{3\pi}{4}$ — the two negative-input rules side by side.

Example 4 — The triangle method: $\sin(\tan^{-1} x)$

Show that $\sin(\tan^{-1} x) = \dfrac{x}{\sqrt{1 + x^2}}$ for $\vert x \vert < 1$.

Step 1 — name the angle: let $\theta = \tan^{-1}x$, so $\tan\theta = x$.

Step 2 — build the right triangle: opposite $x$, adjacent $1$, hypotenuse $\sqrt{1 + x^2}$.

Step 3 — read off the sine: $\sin\theta = \dfrac{x}{\sqrt{1 + x^2}}$ (signs agree on the principal branch).

Answer: $\sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1 + x^2}}$ — the prototype of every triangle-method conversion.

Example 5 — Another triangle read-off

Evaluate $\cos\left(\tan^{-1}\frac{3}{4}\right)$.

Step 1 — the triangle for $\tan\theta = \frac{3}{4}$: opposite 3, adjacent 4, hypotenuse 5.

Step 2 — read off: $\cos\theta = \frac{4}{5}$.

Answer: $\dfrac{4}{5}$.

Example 6 — Tangent of a sum of inverse values

Evaluate $\tan\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right)$.

Step 1 — convert both to tangents: $\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$ (3-4-5 triangle) and $\cot^{-1}\frac{3}{2} = \tan^{-1}\frac{2}{3}$.

Step 2 — tangent addition: $$\tan(A + B) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{6}$$

Answer: $\dfrac{17}{6}$.

Example 7 — Sine of a doubled inverse

Evaluate $\sin\left(2\sin^{-1}\frac{3}{5}\right)$.

Step 1 — name the angle: $\theta = \sin^{-1}\frac{3}{5}$, so $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$.

Step 2 — double-angle formula: $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$.

Answer: $\dfrac{24}{25}$.

Example 8 — A cosine sum identity

Prove that $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$.

Step 1 — name the angles: $A = \cos^{-1}\frac{4}{5}$ (so $\sin A = \frac{3}{5}$), $\ B = \cos^{-1}\frac{12}{13}$ (so $\sin B = \frac{5}{13}$) — both in $\left(0, \frac{\pi}{2}\right)$.

Step 2 — cosine addition: $$\cos(A + B) = \cos A\cos B - \sin A \sin B = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{48 - 15}{65} = \frac{33}{65}$$

Step 3 — branch check: $A + B \in (0, \pi)$, the $\cos^{-1}$ range, so $A + B = \cos^{-1}\frac{33}{65}$. $\blacksquare$

Answer: proved — sum identities always end with a branch check before applying $\cos^{-1}$ to both sides.

Example 9 — A sine sum landing on a tangent

Prove that $\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36}$.

Step 1 — name and complete the triangles: $A = \sin^{-1}\frac{8}{17}$: $\ \cos A = \frac{15}{17}$, $\tan A = \frac{8}{15}$. $\ B = \sin^{-1}\frac{3}{5}$: $\ \tan B = \frac{3}{4}$.

Step 2 — tangent addition: $$\tan(A + B) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15}\cdot\frac{3}{4}} = \frac{\frac{32 + 45}{60}}{\frac{60 - 24}{60}} = \frac{77}{36}$$

Step 3 — branch check: both angles are acute, so $A + B \in (0, \pi)$ with positive tangent, hence $A + B = \tan^{-1}\frac{77}{36}$. $\blacksquare$

Examples 10 to 17 — Sum Identities and Simplifications

Example 10 — Doubling a sine inverse into a tangent

Prove that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$.

Step 1 — name: $\theta = \sin^{-1}\frac{3}{5}$, so $\tan\theta = \frac{3}{4}$.

Step 2 — double-angle for tangent: $$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{24}{7}$$

Step 3 — branch check: $\theta < \frac{\pi}{4}$ (since $\frac{3}{5} < \frac{1}{\sqrt 2}$), so $2\theta \in \left(0, \frac{\pi}{2}\right)$ and $2\theta = \tan^{-1}\frac{24}{7}$. $\blacksquare$

Example 11 — Mixed cosine and sine sum

Prove that $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$.

Step 1 — name: $A = \cos^{-1}\frac{12}{13}$ ($\sin A = \frac{5}{13}$), $\ B = \sin^{-1}\frac{3}{5}$ ($\cos B = \frac{4}{5}$).

Step 2 — sine addition: $$\sin(A + B) = \sin A\cos B + \cos A \sin B = \frac{5}{13}\cdot\frac{4}{5} + \frac{12}{13}\cdot\frac{3}{5} = \frac{20 + 36}{65} = \frac{56}{65}$$

Step 3 — branch check: both acute with sum less than $\frac{\pi}{2}$ (since $\cos(A+B) = \frac{33}{65} > 0$), so the sum equals $\sin^{-1}\frac{56}{65}$. $\blacksquare$

Example 12 — Splitting a tangent inverse

Prove that $\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$.

Step 1 — convert the right side to tangents: $\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}$ and $\cos^{-1}\frac{3}{5} = \tan^{-1}\frac{4}{3}$.

Step 2 — add: $$\tan(A + B) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12}\cdot\frac{4}{3}} = \frac{\frac{5 + 16}{12}}{\frac{36 - 20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{63}{16}$$

Step 3 — branch check: the sum is in $(0, \pi)$ with positive tangent. $\blacksquare$

Example 13 — Root-to-half-angle

Prove that $\tan^{-1}\sqrt x = \dfrac{1}{2}\cos^{-1}\left(\dfrac{1 - x}{1 + x}\right)$ for $x \in [0, 1]$.

Step 1 — substitute $x = \tan^2\theta$: then $$\frac{1 - x}{1 + x} = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$$

Step 2 — apply $\cos^{-1}$: $\ \frac{1}{2}\cos^{-1}(\cos 2\theta) = \frac{1}{2}\cdot 2\theta = \theta = \tan^{-1}\sqrt x$ (branch legal since $2\theta \in [0, \frac{\pi}{2}]$). $\blacksquare$

Example 14 — The nested square roots

Prove that $\cot^{-1}\left(\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right) = \dfrac{x}{2}$ for $x \in \left(0, \frac{\pi}{4}\right)$.

Step 1 — express the roots in half-angles: $$1 \pm \sin x = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2 \implies \sqrt{1 + \sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}, \quad \sqrt{1 - \sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}$$ (the window makes both parenthesised quantities positive, so the moduli drop).

Step 2 — simplify the big quotient: $$\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}$$

Step 3 — cancel: $\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}$. $\blacksquare$

Example 15 — The $\sqrt{1+x}, \sqrt{1-x}$ quotient

Prove that $\tan^{-1}\left(\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}\right) = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}x$ for $-\frac{1}{\sqrt 2} \leq x \leq 1$.

Step 1 — substitute $x = \cos 2\theta$: then $\sqrt{1 + x} = \sqrt 2\cos\theta$ and $\sqrt{1 - x} = \sqrt 2 \sin\theta$.

Step 2 — the quotient becomes: $$\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$$

Step 3 — cancel and back-substitute $\theta = \frac{1}{2}\cos^{-1}x$: $$\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x \qquad \blacksquare$$

Example 16 — The $\sqrt{1 + x^2} - 1$ trick

Write $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)$, $x \neq 0$, in simplest form.

Step 1 — substitute $x = \tan\theta$: then $\sqrt{1 + x^2} = \sec\theta$ and $$\frac{\sec\theta - 1}{\tan\theta} = \frac{1 - \cos\theta}{\sin\theta} = \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}$$

Step 2 — cancel: the expression equals $\frac{\theta}{2} = \frac{1}{2}\tan^{-1}x$.

Answer: $\dfrac{1}{2}\tan^{-1} x$ — the half-angle identities convert the awkward surd into a clean half.

Example 17 — The triple-angle tangent

Write $\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right)$, $\ a > 0$, $-\frac{a}{\sqrt 3} < x < \frac{a}{\sqrt 3}$, in simplest form.

Step 1 — substitute $x = a\tan\theta$: numerator and denominator become $a^3(3\tan\theta - \tan^3\theta)$ and $a^3(1 - 3\tan^2\theta)$.

Step 2 — recognise $\tan 3\theta$: $$\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta$$

Step 3 — cancel (the window keeps $3\theta$ inside the branch):

Answer: $3\tan^{-1}\dfrac{x}{a}$.

Examples 18 to 25 — Equations and Closing Classics

Example 18 — An inverse-trig equation

Solve $2\tan^{-1}(\cos x) = \tan^{-1}(2\,\mathrm{cosec}\,x)$.

Step 1 — apply the double formula on the left (with $t = \cos x$): $$2\tan^{-1}t = \tan^{-1}\frac{2t}{1 - t^2} \implies \tan^{-1}\frac{2\cos x}{1 - \cos^2 x} = \tan^{-1}\frac{2\cos x}{\sin^2 x}$$

Step 2 — equate the arguments: $$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x} \implies \frac{\cos x}{\sin x} = 1 \implies \tan x = 1$$

Answer: $x = \dfrac{\pi}{4}$ (taking the principal solution, which also validates the formula used in Step 1).

Example 19 — A half-tangent equation

Solve $\tan^{-1}\dfrac{1 - x}{1 + x} = \dfrac{1}{2}\tan^{-1}x$, $\ x > 0$.

Step 1 — recognise the left side: $\dfrac{1 - x}{1 + x} = \tan\left(\frac{\pi}{4} - \tan^{-1}x\right)$, so the equation reads $$\frac{\pi}{4} - \tan^{-1}x = \frac{1}{2}\tan^{-1}x$$

Step 2 — solve for $\tan^{-1}x$: $\ \frac{3}{2}\tan^{-1}x = \frac{\pi}{4}$, so $\tan^{-1}x = \frac{\pi}{6}$.

Answer: $x = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt 3}$.

Example 20 — The equation with an extraneous root

Solve $\sin^{-1}(1 - x) - 2\sin^{-1}x = \dfrac{\pi}{2}$.

Step 1 — isolate and take sine: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x$, so $$1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right) = \cos(2\sin^{-1}x) = 1 - 2x^2$$

Step 2 — solve the quadratic: $2x^2 - x = 0$ gives $x = 0$ or $x = \frac{1}{2}$.

Step 3 — CHECK both in the original (taking sine is not reversible): $x = 0$: $\ \sin^{-1}1 - 0 = \frac{\pi}{2}$ ✓. $\ x = \frac{1}{2}$: $\ \sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = -\frac{\pi}{6} \neq \frac{\pi}{2}$ ✗.

Answer: $x = 0$ only — the quadratic's second root is extraneous. Always substitute back in inverse-trig equations.

Example 21 — The double-angle forms combined

Prove that $\tan\left(\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1 + x^2} + \cos^{-1}\dfrac{1 - y^2}{1 + y^2}\right]\right) = \dfrac{x + y}{1 - xy}$, for $\vert x \vert < 1$, $y > 0$, $xy < 1$.

Step 1 — recognise both double-angle disguises: $$\sin^{-1}\frac{2x}{1 + x^2} = 2\tan^{-1}x, \qquad \cos^{-1}\frac{1 - y^2}{1 + y^2} = 2\tan^{-1}y$$ (the stated windows make both legal).

Step 2 — the half kills the 2's: $$\tan\left(\tan^{-1}x + \tan^{-1}y\right) = \frac{x + y}{1 - xy} \qquad \blacksquare$$

Example 22 — Cosine of a sum of two sine inverses

Evaluate $\cos\left(\sin^{-1}\dfrac{3}{5} + \sin^{-1}\dfrac{5}{13}\right)$.

Step 1 — complete both triangles: $\sin A = \frac{3}{5}, \cos A = \frac{4}{5}$; $\ \sin B = \frac{5}{13}, \cos B = \frac{12}{13}$.

Step 2 — cosine addition: $$\cos(A + B) = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{48 - 15}{65} = \frac{33}{65}$$

Answer: $\dfrac{33}{65}$.

Example 23 — An extreme-value argument

If $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \dfrac{3\pi}{2}$, find $x + y + z$.

Step 1 — bound each term: every $\sin^{-1}$ value is at most $\frac{\pi}{2}$, so the sum of three is at most $\frac{3\pi}{2}$.

Step 2 — equality forces the maximum everywhere: each term must equal $\frac{\pi}{2}$, i.e. $x = y = z = 1$.

Answer: $x + y + z = 3$. (Extreme-value forcing is a standard JEE finisher: when a sum hits its theoretical maximum, every summand is pinned.)

Example 24 — Domain hunting

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

Step 1 — the inner expression must lie in $[-1, 1]$: $$-1 \leq 2x - 1 \leq 1$$

Step 2 — solve: $0 \leq 2x \leq 2$, so $0 \leq x \leq 1$.

Answer: $[0, 1]$ — domain questions for inverse trig are pure inequality-solving against the branch table.

Example 25 — A two-root equation, both valid

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

Step 1 — move and take sine: rewrite as $\sin^{-1}(1 - x) = \cos^{-1}x - \sin^{-1}x$ and take sine of both sides, using $\sin(\cos^{-1}x) = \cos(\sin^{-1}x) = \sqrt{1 - x^2}$: $$1 - x = \sin(\cos^{-1}x)\cos(\sin^{-1}x) - \cos(\cos^{-1}x)\sin(\sin^{-1}x) = \left(\sqrt{1 - x^2}\right)^2 - x \cdot x = 1 - 2x^2$$

Step 2 — solve and check: $1 - x = 1 - 2x^2$ gives $2x^2 - x = 0$, i.e. $x = 0$ or $x = \frac{1}{2}$. Substitute both back: At $x = 0$: $\ \sin^{-1}0 + \sin^{-1}1 = \frac{\pi}{2} = \cos^{-1}0$ ✓. At $x = \frac{1}{2}$: $\ \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} = \cos^{-1}\frac{1}{2}$ ✓.

Answer: $x = 0$ or $x = \dfrac{1}{2}$ — this time both roots survive. Compare with Example 20: the identical quadratic $2x^2 - x = 0$ arose there too, but with a different original equation only one root survived. Substitution back is what separates them.