Inverse trigonometric functions

arcsin, arccos, arctan and arccot: values, compositions, domains and equations. 8 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 Basic values

Find each value in radians.

a) \(\arcsin\frac12\) b) \(\arcsin\left(-\frac{\sqrt2}{2}\right)\) c) \(\arccos\left(-\frac12\right)\) d) \(\arccos\frac{\sqrt3}{2}\) e) \(\arctan(-1)\)

f) \(\arctan\frac{\sqrt3}{3}\) g) \(\operatorname{arccot}(-\sqrt3)\) h) \(\operatorname{arccot} 1\) i) \(\arcsin(-1)\) j) \(\arccos 0\)

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Ranges: \(\arcsin \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), \(\arccos \in [0, \pi]\), \(\arctan \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \(\operatorname{arccot} \in (0, \pi)\).

a) \(\frac{\pi}{6}\) b) \(-\frac{\pi}{4}\) c) \(\frac{2\pi}{3}\) d) \(\frac{\pi}{6}\) e) \(-\frac{\pi}{4}\)
f) \(\frac{\pi}{6}\) g) \(\frac{5\pi}{6}\) h) \(\frac{\pi}{4}\) i) \(-\frac{\pi}{2}\) j) \(\frac{\pi}{2}\)

Problem 2 Function and inverse

Find.

a) \(\sin\left(\arcsin\frac37\right)\) b) \(\cos\left(\arccos\left(-\frac25\right)\right)\) c) \(\arcsin\left(\sin\frac{\pi}{5}\right)\) d) \(\arcsin\left(\sin\frac{4\pi}{5}\right)\) e) \(\arccos\left(\cos\left(-\frac{\pi}{3}\right)\right)\) f) \(\arctan\left(\tan\frac{7\pi}{6}\right)\)

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a) \(\frac37\) and b) \(-\frac25\): for \(a\) in the domain, \(\sin(\arcsin a) = a\) and \(\cos(\arccos a) = a\).
c) \(\frac{\pi}{5}\) is in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so the result is \(\frac{\pi}{5}\).
d) \(\sin\frac{4\pi}{5} = \sin\frac{\pi}{5}\), so the result is \(\frac{\pi}{5}\) (not \(\frac{4\pi}{5}\)).
e) \(\cos\left(-\frac{\pi}{3}\right) = \frac12\), so the result is \(\arccos\frac12 = \frac{\pi}{3}\).
f) \(\tan\frac{7\pi}{6} = \tan\frac{\pi}{6}\), so the result is \(\frac{\pi}{6}\).

Problem 3 Sums of inverse values

Calculate.

a) \(\arccos\frac{\sqrt2}{2} + \arcsin\frac{\sqrt3}{2}\) b) \(\arctan 1 + 2\arctan\sqrt3 - \arctan\left(-\frac{\sqrt3}{3}\right)\) c) \(\arccos\left(-\frac{\sqrt3}{2}\right) - \arcsin\frac12\)

d) If \(a = \arcsin\frac{\sqrt3}{2}\) and \(b = \arcsin\left(-\frac{\sqrt2}{2}\right)\), find \(a + b\).

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a) \(\frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}\)
b) \(\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{6} = \frac{3\pi + 8\pi + 2\pi}{12} = \frac{13\pi}{12}\)
c) \(\frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}\)
d) \(\frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}\)

Problem 4 Compositions

Evaluate.

a) \(\cos\left(\arcsin\frac45\right)\) b) \(\tan\left(\arcsin\frac13\right)\) c) \(\sin(\arctan 3)\) d) \(\cot(\arctan 4)\) e) \(\cos\left(\arctan\frac12\right)\) f) \(\sin\left(\arccos\left(-\frac35\right)\right)\) g) \(\tan\left(\operatorname{arccot}\frac13\right)\)

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Put \(\alpha\) = the inner angle and use a right triangle (or \(\sin^2 + \cos^2 = 1\)), remembering the sign from the range.

a) \(\sin\alpha = \frac45\), \(\alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so \(\cos\alpha = \frac35\).
b) \(\sin\alpha = \frac13\), \(\cos\alpha = \frac{2\sqrt2}{3}\), \(\tan\alpha = \frac{1}{2\sqrt2} = \frac{\sqrt2}{4}\).
c) \(\tan\alpha = 3\): opposite 3, adjacent 1, hypotenuse \(\sqrt{10}\): \(\sin\alpha = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}\).
d) \(\frac14\).
e) opposite 1, adjacent 2, hypotenuse \(\sqrt5\): \(\cos\alpha = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\).
f) \(\alpha \in [0, \pi]\), so \(\sin\alpha \ge 0\): \(\sin\alpha = \sqrt{1 - \frac{9}{25}} = \frac45\).
g) \(\cot\alpha = \frac13\), so \(\tan\alpha = 3\).

Problem 5 Double angles and reductions

Evaluate.

a) \(\sin\left(2\arcsin\frac13\right)\) b) \(\cos\left(2\arcsin\frac35\right)\) c) \(\tan\left(2\arctan\frac13\right)\) d) \(\sin\left(\pi - \arcsin\frac25\right)\) e) \(\sin\left(\pi + \arcsin\frac14\right)\)

f) \(\cos\left(\frac{\pi}{2} - \arcsin\frac23\right)\) g) \(\tan\left(\frac{3\pi}{2} + \arctan\frac25\right)\) h) \(\cot\left(\frac{\pi}{2} - \arcsin\frac12\right)\)

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a) \(2\sin\alpha\cos\alpha = 2\cdot\frac13\cdot\frac{2\sqrt2}{3} = \frac{4\sqrt2}{9}\)
b) \(1 - 2\sin^2\alpha = 1 - \frac{18}{25} = \frac{7}{25}\)
c) \(\frac{2\cdot\frac13}{1 - \frac19} = \frac{2/3}{8/9} = \frac34\)
d) \(\sin(\pi - \alpha) = \sin\alpha = \frac25\)
e) \(\sin(\pi + \alpha) = -\sin\alpha = -\frac14\)
f) \(\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha = \frac23\)
g) \(\tan\left(\frac{3\pi}{2} + \alpha\right) = -\cot\alpha = -\frac52\)
h) \(\tan\left(\arcsin\frac12\right) = \tan\frac{\pi}{6} = \frac{\sqrt3}{3}\)

Problem 6 Mixed expressions

Evaluate.

a) \(\sin\left(\arcsin\frac{\sqrt2}{2} + \arccos\frac12\right)\) b) \(\arcsin\left[\cos\left(\arcsin\frac12\right)\right]\) c) \(\cos(\arcsin(\cos 60^\circ))\)

d) \(\arcsin(\tan 45^\circ)\) e) \(\arccos(\cos(-60^\circ))\) f) \(\tan(\operatorname{arccot}\sqrt3) + \cos(\operatorname{arccot} 1) - \sin(\operatorname{arccot} 0)\)

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a) \(\sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin\frac{7\pi}{12} = \frac{\sqrt6 + \sqrt2}{4}\)
b) \(\cos\frac{\pi}{6} = \frac{\sqrt3}{2}\), so we get \(\arcsin\frac{\sqrt3}{2} = \frac{\pi}{3}\)
c) \(\cos\left(\arcsin\frac12\right) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2}\)
d) \(\arcsin 1 = \frac{\pi}{2}\)
e) \(\arccos\frac12 = \frac{\pi}{3}\) (that is \(60^\circ\))
f) \(\tan\frac{\pi}{6} + \cos\frac{\pi}{4} - \sin\frac{\pi}{2} = \frac{\sqrt3}{3} + \frac{\sqrt2}{2} - 1\)

Problem 7 Domain

a) State the domain: (i) \(f(x) = \arcsin(2x - 1)\) (ii) \(f(x) = \arccos(3x + 2)\) (iii) \(f(x) = \arcsin\frac{x}{4}\) (iv) \(f(x) = \arccos(1 - 2x)\).

b) \(f(x) = \arccos(2x - 1)\). Find \(f(0)\) and \(f\left(\frac14\right)\).

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a) The argument must be in \([-1, 1]\).

(i) \(-1 \le 2x - 1 \le 1 \Rightarrow 0 \le x \le 1\): \([0, 1]\)
(ii) \(-1 \le 3x + 2 \le 1 \Rightarrow -1 \le x \le -\frac13\): \(\left[-1, -\frac13\right]\)
(iii) \([-4, 4]\)
(iv) \(-1 \le 1 - 2x \le 1 \Rightarrow 0 \le x \le 1\): \([0, 1]\)

b) \(f(0) = \arccos(-1) = \pi\); \(f\left(\frac14\right) = \arccos\left(-\frac12\right) = \frac{2\pi}{3}\).

Problem 8 Equations with inverse functions

Solve.

a) \(\arcsin x - \arccos x = \frac{\pi}{6}\) b) \(\arctan(x + 1) - \arctan(x - 1) = \frac{\pi}{4}\)

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a) Use \(\arccos x = \frac{\pi}{2} - \arcsin x\): \(2\arcsin x - \frac{\pi}{2} = \frac{\pi}{6}\), so \(\arcsin x = \frac{\pi}{3}\) and \(x = \frac{\sqrt3}{2}\).

b) Take the tangent of both sides: \(\frac{(x + 1) - (x - 1)}{1 + (x + 1)(x - 1)} = 1\), so \(\frac{2}{x^2} = 1\) and \(x = \pm\sqrt2\).

Check: for \(x = \sqrt2\): \(\arctan(\sqrt2 + 1) - \arctan(\sqrt2 - 1) = 67.5^\circ - 22.5^\circ = 45^\circ\). For \(x = -\sqrt2\): \(\arctan(1 - \sqrt2) - \arctan(-1 - \sqrt2) = -22.5^\circ + 67.5^\circ = 45^\circ\). Both are solutions.