The Cancellation Laws and Their Traps

Two laws, two very different validity windows

Because sin⁡−1\sin^{-1} inverts the restricted sine, the two cancellation identities hold on different sets:

sin⁡(sin⁡−1x)=xfor x∈[−1,1](always works on the domain)\sin(\sin^{-1} x) = x \quad \text{for } x \in [-1, 1] \qquad \text{(always works on the domain)} sin⁡−1(sin⁡x)=xonly for x∈[−π2,π2](the branch window)\sin^{-1}(\sin x) = x \quad \text{only for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \qquad \text{(the branch window)}

The first is safe: feed any legal xx into sin⁡−1\sin^{-1}, take sine, get xx back. The second is the trap: sin⁡−1(sin⁡x)\sin^{-1}(\sin x) returns not xx but the angle in the principal branch with the same sine as xx.

The wrap-around technique

Problem. Find sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right).

Step 1 — notice the input is outside the branch: 3π5∉[−π2,π2]\frac{3\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so the answer is NOT 3π5\frac{3\pi}{5}.

Step 2 — rewrite with the same sine, inside the branch: sin⁡3π5=sin⁡(π−3π5)=sin⁡2π5,2π5∈[−π2,π2]✓\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}, \qquad \frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ✓

Step 3 — now cancel: sin⁡−1(sin⁡3π5)=2π5\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \frac{2\pi}{5}.

The identities to wrap with: sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x,  cos⁡(2π−x)=cos⁡x\ \cos(2\pi - x) = \cos x,  tan⁡(x−π)=tan⁡x\ \tan(x - \pi) = \tan x — choose whichever lands the angle in the principal range.

More worked wrap-arounds: cos⁡−1(cos⁡7π6)=cos⁡−1(cos⁡5π6)=5π6(cos⁡7π6=cos⁡(2π−7π6))\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \cos^{-1}\left(\cos\frac{5\pi}{6}\right) = \frac{5\pi}{6} \quad \left(\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right)\right) tan⁡−1(tan⁡3π4)=tan⁡−1(tan⁡(3π4−π))=−π4\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}\left(\tan\left(\frac{3\pi}{4} - \pi\right)\right) = -\frac{\pi}{4} sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡π3)=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3}

The triple-angle pair (proved by substitution)

3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3cos⁡−1x=cos⁡−1(4x3−3x),x∈[12,1]3\sin^{-1} x = \sin^{-1}(3x - 4x^3), \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \qquad \qquad 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \quad x \in \left[\frac{1}{2}, 1\right]

Proof of the first: put x=sin⁡θx = \sin\theta. Then sin⁡−1(3x−4x3)=sin⁡−1(3sin⁡θ−4sin⁡3θ)=sin⁡−1(sin⁡3θ)=3θ=3sin⁡−1x\sin^{-1}(3x - 4x^3) = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x — the restriction on xx is exactly what keeps 3θ3\theta inside the principal branch, so the cancellation is legal. The second is identical with x=cos⁡θx = \cos\theta and cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta.

Common mistakes to avoid

Mistake 1 — cancelling blindly: sin⁡−1(sin⁡3π5)≠3π5\sin^{-1}\left(\sin\frac{3\pi}{5}\right) \neq \frac{3\pi}{5}. Always ask: is the angle inside the principal branch?

Mistake 2 — wrapping with the wrong identity: for cos⁡−1\cos^{-1} use cos⁡(2π−x)\cos(2\pi - x) or cos⁡(−x)\cos(-x) (both preserve cosine); using cos⁡(π−x)=−cos⁡x\cos(\pi - x) = -\cos x changes the sign.

Mistake 3 — ignoring the restriction in the triple-angle formulas: for xx outside [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right],  sin⁡−1(3x−4x3)\ \sin^{-1}(3x - 4x^3) no longer equals 3sin⁡−1x3\sin^{-1}x (the left side is trapped in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] while 3sin⁡−1x3\sin^{-1}x escapes it).

Simplest Forms by Substitution

The substitution dictionary

Most "write in simplest form" problems dissolve under one well-chosen substitution. Match the algebraic shape to the trig identity it hides:

See this shape Substitute Because
1−x2\sqrt{1 - x^2} x=sin⁡θx = \sin\theta or x=cos⁡θx = \cos\theta 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta
1+x2\sqrt{1 + x^2} x=tan⁡θx = \tan\theta 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
x2−1\sqrt{x^2 - 1} x=sec⁡θx = \sec\theta sec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta
a2−x2\sqrt{a^2 - x^2} x=asin⁡θx = a\sin\theta scaled version
1−cos⁡x1 - \cos x, 1+cos⁡x1 + \cos x half angles 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\frac{x}{2},  1+cos⁡x=2cos⁡2x2\ 1 + \cos x = 2\cos^2\frac{x}{2}
cos⁡x±sin⁡x\cos x \pm \sin x factor 2\sqrt 2 or divide by cos⁡x\cos x builds tan⁡(π4±x)\tan\left(\frac{\pi}{4} \pm x\right)

Three model reductions

Model 1.  sin⁡−1(2x1−x2)\ \sin^{-1}\left(2x\sqrt{1 - x^2}\right),  −12≤x≤12\ -\frac{1}{\sqrt 2} \leq x \leq \frac{1}{\sqrt 2}.

Put x=sin⁡θx = \sin\theta: the expression becomes sin⁡−1(2sin⁡θcos⁡θ)=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1x\sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin 2\theta) = 2\theta = \mathbf{2\sin^{-1} x}. (For 12≤x≤1\frac{1}{\sqrt 2} \leq x \leq 1, put x=cos⁡θx = \cos\theta instead and the same expression equals 2cos⁡−1x2\cos^{-1}x — the substitution must respect the branch.)

Model 2.  tan⁡−1cos⁡x1−sin⁡x\ \tan^{-1}\dfrac{\cos x}{1 - \sin x},  −3π2<x<π2\ -\frac{3\pi}{2} < x < \frac{\pi}{2}.

Write numerator and denominator in half-angles: cos⁡x1−sin⁡x=cos⁡2x2−sin⁡2x2(cos⁡x2−sin⁡x2)2=cos⁡x2+sin⁡x2cos⁡x2−sin⁡x2=1+tan⁡x21−tan⁡x2=tan⁡(π4+x2)\frac{\cos x}{1 - \sin x} = \frac{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) so the whole expression is π4+x2\mathbf{\dfrac{\pi}{4} + \dfrac{x}{2}}.

Model 3.  cot⁡−11x2−1\ \cot^{-1}\dfrac{1}{\sqrt{x^2 - 1}},  x>1\ x > 1.

Put x=sec⁡θx = \sec\theta: then x2−1=tan⁡θ\sqrt{x^2 - 1} = \tan\theta, and cot⁡−1(1tan⁡θ)=cot⁡−1(cot⁡θ)=θ=sec⁡−1x\cot^{-1}\left(\frac{1}{\tan\theta}\right) = \cot^{-1}(\cot\theta) = \theta = \mathbf{\sec^{-1} x}.

The standard results worth memorising

tan⁡−11+x2−1x=12tan⁡−1xtan⁡−11−cos⁡x1+cos⁡x=x2 (0<x<π)\tan^{-1}\frac{\sqrt{1 + x^2} - 1}{x} = \frac{1}{2}\tan^{-1}x \qquad \tan^{-1}\sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{x}{2} \ (0 < x < \pi) tan⁡−1cos⁡x−sin⁡xcos⁡x+sin⁡x=π4−xtan⁡−1xa2−x2=sin⁡−1xa\tan^{-1}\frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\pi}{4} - x \qquad \tan^{-1}\frac{x}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} tan⁡−13a2x−x3a3−3ax2=3tan⁡−1xa (−a3<x<a3)\tan^{-1}\frac{3a^2x - x^3}{a^3 - 3ax^2} = 3\tan^{-1}\frac{x}{a} \ \left(-\frac{a}{\sqrt 3} < x < \frac{a}{\sqrt 3}\right)

Each comes from the dictionary: the first from x=tan⁡θx = \tan\theta with half-angle collapse, the second and third from half-angle identities, the fourth from x=asin⁡θx = a\sin\theta, the fifth from x=atan⁡θx = a\tan\theta and the tan⁡3θ\tan 3\theta formula.

Common mistakes to avoid

Mistake 1 — forgetting the validity interval. Each reduction is stated with a window (like x>1x > 1 or 0<x<π0 < x < \pi); outside it, signs and branches shift and the clean answer fails.

Mistake 2 — cancelling (⋅)2\sqrt{(\cdot)^2} carelessly: sin⁡2x2=∣sin⁡x2∣\sqrt{\sin^2\frac{x}{2}} = \left\vert \sin\frac{x}{2} \right\vert — the given window is what lets the modulus drop.

Mistake 3 — substituting against the shape: putting x=tan⁡θx = \tan\theta into 1−x2\sqrt{1 - x^2} creates a mess; the dictionary pairing is not optional.

Solved Examples

Example 1 — A wrap-around sine

Find sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right).

Step 1 — branch check: 3π5>π2\frac{3\pi}{5} > \frac{\pi}{2}, so direct cancellation is illegal.

Step 2 — wrap: sin⁡3π5=sin⁡(π−3π5)=sin⁡2π5\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}, and 2π5∈[−π2,π2]\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Answer: 2π5\dfrac{2\pi}{5}.

Example 2 — A wrap-around cosine

Find cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\frac{7\pi}{6}\right).

Step 1 — branch check: 7π6>π\frac{7\pi}{6} > \pi, outside [0,π][0, \pi].

Step 2 — wrap with cos⁡(2π−x)=cos⁡x\cos(2\pi - x) = \cos x:  cos⁡7π6=cos⁡(2π−7π6)=cos⁡5π6\ \cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}, and 5π6∈[0,π]\frac{5\pi}{6} \in [0, \pi] ✓.

Answer: 5π6\dfrac{5\pi}{6} — not 7π6\frac{7\pi}{6}.

Example 3 — A wrap-around tangent

Find tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\frac{3\pi}{4}\right).

Step 1 — branch check: 3π4∉(−π2,π2)\frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Step 2 — wrap with period π\pi: tan⁡3π4=tan⁡(3π4−π)=tan⁡(−π4)\tan\frac{3\pi}{4} = \tan\left(\frac{3\pi}{4} - \pi\right) = \tan\left(-\frac{\pi}{4}\right).

Answer: −π4-\dfrac{\pi}{4}.

Example 4 — The double-angle reduction, both branches

Show that (i) sin⁡−1(2x1−x2)=2sin⁡−1x\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1}x for −12≤x≤12-\frac{1}{\sqrt 2} \leq x \leq \frac{1}{\sqrt 2}, and (ii) sin⁡−1(2x1−x2)=2cos⁡−1x\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\cos^{-1}x for 12≤x≤1\frac{1}{\sqrt 2} \leq x \leq 1.

Step 1 — (i) substitute x=sin⁡θx = \sin\theta: then 1−x2=cos⁡θ\sqrt{1 - x^2} = \cos\theta and sin⁡−1(2sin⁡θcos⁡θ)=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1x\sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x legal because the given window keeps 2θ∈[−π2,π2]2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Step 2 — (ii) substitute x=cos⁡θx = \cos\theta: the same expression becomes sin⁡−1(sin⁡2θ)=2θ=2cos⁡−1x\sin^{-1}(\sin 2\theta) = 2\theta = 2\cos^{-1}x, legal on the new window.

Answer: the same expression simplifies differently on different intervals — the substitution must match the range of xx.

Example 5 — The half-angle quotient

Express tan⁡−1cos⁡x1−sin⁡x\tan^{-1}\dfrac{\cos x}{1 - \sin x},  −3π2<x<π2\ -\frac{3\pi}{2} < x < \frac{\pi}{2}, in simplest form.

Step 1 — half-angle rewrite: cos⁡x=cos⁡2x2−sin⁡2x2\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} and 1−sin⁡x=(cos⁡x2−sin⁡x2)21 - \sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2.

Step 2 — cancel the common factor: (cos⁡x2+sin⁡x2)(cos⁡x2−sin⁡x2)(cos⁡x2−sin⁡x2)2=1+tan⁡x21−tan⁡x2=tan⁡(π4+x2)\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)

Step 3 — cancel with tan⁡−1\tan^{-1}:

Answer: π4+x2\dfrac{\pi}{4} + \dfrac{x}{2}.

Example 6 — The secant substitution

Write cot⁡−11x2−1\cot^{-1}\dfrac{1}{\sqrt{x^2 - 1}},  x>1\ x > 1, in simplest form.

Step 1 — the shape x2−1\sqrt{x^2 - 1} calls for x=sec⁡θx = \sec\theta: then x2−1=tan⁡θ\sqrt{x^2 - 1} = \tan\theta (positive here).

Step 2 — reduce: cot⁡−1(1tan⁡θ)=cot⁡−1(cot⁡θ)=θ\cot^{-1}\left(\frac{1}{\tan\theta}\right) = \cot^{-1}(\cot\theta) = \theta.

Answer: sec⁡−1x\sec^{-1} x.

Example 7 — A nested evaluation

Find tan⁡−1(2cos⁡(2sin⁡−112))\tan^{-1}\left(2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right).

Step 1 — innermost first: sin⁡−112=π6\sin^{-1}\frac{1}{2} = \frac{\pi}{6}.

Step 2 — work outwards: 2cos⁡(π3)=2⋅12=12\cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1.

Step 3 — outermost: tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}.

Answer: π4\dfrac{\pi}{4} — nested expressions unwind strictly from the inside out.