Which angle do we mean?

sin⁡x=12\displaystyle \sin x = \tfrac{1}{2} has infi­nitely many solu­tions. So if some­one asks for "the angle whose sine is 12\displaystyle \tfrac{1}{2}", we need an agreed rule to choose just one. The rule is to restrict each trigono­met­ric func­tion to a stretch, a branch, where it takes every value exactly once. On that branch it can be undone, and that gives us the inverse trigono­met­ric func­tions. We will see their prin­ci­pal-value ranges and graphs, and prac­tise find­ing prin­ci­pal val­ues.

The prin­ci­pal branches

Func­tion Domain Range (prin­ci­pal value)
sin⁡−1x\sin^{-1}x [−1,1][-1, 1] [−π2,π2]\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]
cos⁡−1x\cos^{-1}x [−1,1][-1, 1] [0,π][0, \pi]
tan⁡−1x\tan^{-1}x R\mathbf{R} (−π2,π2)\displaystyle \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)
cot⁡−1x\cot^{-1}x R\mathbf{R} (0,π)(0, \pi)
sec⁡−1x\sec^{-1}x R−(−1,1)\mathbf{R} - (-1, 1) [0,π]−{π2}\displaystyle [0, \pi] - \left\{\tfrac{\pi}{2}\right\}
csc⁡−1x\csc^{-1}x R−(−1,1)\mathbf{R} - (-1, 1) [−π2,π2]−{0}\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}

A com­mon slip to avoid: sin⁡−1x\sin^{-1}x is not 1sin⁡x\displaystyle \tfrac{1}{\sin x}. The graph of sin⁡−1\sin^{-1} is the graph of sin⁡\sin on [−π2,π2]\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] reflected in the line y=xy = x.

The graph of y = sin x for x from -pi/2 to pi/2 and its mirror image y = sin inverse x, from -1 to 1, reflected in the dashed line y = x.
The graph of inverse sine is the restricted sine graph reflected in y = x.

Prin­ci­pal val­ues

Exam­ple 1. sin⁡−1(−32)=−π3\displaystyle \sin^{-1}\left(-\tfrac{\sqrt{3}}{2}\right) = -\tfrac{\pi}{3} (it lies in the range, and sin⁡(−π3)=−32\displaystyle \sin(-\tfrac{\pi}{3}) = -\tfrac{\sqrt{3}}{2}).

Exam­ple 2. cos⁡−1(−12)=3π4\displaystyle \cos^{-1}\left(-\tfrac{1}{\sqrt{2}}\right) = \tfrac{3\pi}{4}. It is not −π4\displaystyle -\tfrac{\pi}{4}, because that lies out­side [0,π][0, \pi].

Graph of y = cos inverse x for x from -1 to 1, with values between 0 and pi; the point with x = -1/root 2 is marked at height 3 pi/4.
Inverse cosine takes val­ues only in [0, pi], so the value at -1/root 2 is 3pi/4.

Exam­ple 3. tan⁡−1(−3)=−π3\displaystyle \tan^{-1}(-\sqrt{3}) = -\tfrac{\pi}{3}; cot⁡−1(−1)=3π4\displaystyle \cot^{-1}(-1) = \tfrac{3\pi}{4}; sec⁡−1(−2)=2π3\displaystyle \sec^{-1}(-2) = \tfrac{2\pi}{3}; csc⁡−1(2)=π4\displaystyle \csc^{-1}(\sqrt{2}) = \tfrac{\pi}{4}.

Exam­ple 4. tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)=π4+2π3−π6=3π4\displaystyle \tan^{-1}(1) + \cos^{-1}\left(-\tfrac{1}{2}\right) + \sin^{-1}\left(-\tfrac{1}{2}\right) = \tfrac{\pi}{4} + \tfrac{2\pi}{3} - \tfrac{\pi}{6} = \tfrac{3\pi}{4}.

Watch the range. This is where most marks are lost. sin⁡−1(sin⁡5π6)=π6\displaystyle \sin^{-1}\left(\sin \tfrac{5\pi}{6}\right) = \tfrac{\pi}{6}, not 5π6\displaystyle \tfrac{5\pi}{6}. Why? Because sin⁡5π6=12\displaystyle \sin\tfrac{5\pi}{6} = \tfrac{1}{2}, and the prin­ci­pal value for that is π6\displaystyle \tfrac{\pi}{6}. In the same way, cos⁡−1(cos⁡7π4)=π4\displaystyle \cos^{-1}\left(\cos\tfrac{7\pi}{4}\right) = \tfrac{\pi}{4} and tan⁡−1(tan⁡4π5)=−π5\displaystyle \tan^{-1}\left(\tan\tfrac{4\pi}{5}\right) = -\tfrac{\pi}{5}.

Try these your­self

  1. Find the prin­ci­pal val­ues: sin⁡−112\displaystyle \sin^{-1}\tfrac{1}{\sqrt{2}}; cos⁡−132\displaystyle \cos^{-1}\tfrac{\sqrt{3}}{2}; tan⁡−113\displaystyle \tan^{-1}\tfrac{1}{\sqrt{3}}; csc⁡−1(−2)\csc^{-1}(-2); sec⁡−123\displaystyle \sec^{-1}\tfrac{2}{\sqrt{3}}; cot⁡−1(−3)\cot^{-1}(-\sqrt{3}).
  2. Find cos⁡−1(−32)+sin⁡−1(−1)\displaystyle \cos^{-1}\left(-\tfrac{\sqrt{3}}{2}\right) + \sin^{-1}(-1).
  3. Find tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt{3} - \sec^{-1}(-\sqrt{2}).
  4. Find sin⁡−1(sin⁡3π4)\displaystyle \sin^{-1}\left(\sin\tfrac{3\pi}{4}\right), cos⁡−1(cos⁡5π3)\displaystyle \cos^{-1}\left(\cos\tfrac{5\pi}{3}\right), tan⁡−1(tan⁡7π6)\displaystyle \tan^{-1}\left(\tan\tfrac{7\pi}{6}\right).
  5. State the domain and range of cos⁡−1x\cos^{-1}x and sketch its graph.
  6. Find the domain of sin⁡−1(2x−3)\sin^{-1}(2x - 3).
  7. Is sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for all real xx? For which xx?
  8. Eval­u­ate sin⁡(π3−sin⁡−1(−12))\displaystyle \sin\left(\tfrac{\pi}{3} - \sin^{-1}\left(-\tfrac{1}{2}\right)\right).

Answers to check against

Show answers
  1. π4\displaystyle \tfrac{\pi}{4}; π6\displaystyle \tfrac{\pi}{6}; π6\displaystyle \tfrac{\pi}{6}; −π6\displaystyle -\tfrac{\pi}{6}; π6\displaystyle \tfrac{\pi}{6}; 5π6\displaystyle \tfrac{5\pi}{6}.
  2. 5π6−π2=π3\displaystyle \tfrac{5\pi}{6} - \tfrac{\pi}{2} = \tfrac{\pi}{3}.
  3. π3−3π4=−5π12\displaystyle \tfrac{\pi}{3} - \tfrac{3\pi}{4} = -\tfrac{5\pi}{12}.
  4. π4\displaystyle \tfrac{\pi}{4}; π3\displaystyle \tfrac{\pi}{3}; π6\displaystyle \tfrac{\pi}{6}.
  5. [−1,1][-1, 1]; [0,π][0, \pi]; the graph is a decreas­ing curve from (−1,π)(-1, \pi) to (1,0)(1, 0).
  6. −1≤2x−3≤1-1 \le 2x - 3 \le 1: [1,2][1, 2].
  7. No. It holds only for x∈[−π2,π2]\displaystyle x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right].
  8. sin⁡(π3+π6)=1\displaystyle \sin\left(\tfrac{\pi}{3} + \tfrac{\pi}{6}\right) = 1.