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Concept 10 / 11Trigonometry

Inverse Trigonometric Functions: Principal Values, Domains, Ranges and Graphs

Why the trigonometric functions must be restricted to be inverted, the principal value branches of sin^-1, cos^-1, tan^-1, cot^-1, sec^-1 and csc^-1, their graphs, the difference between sin(sin^-1 x) and sin^-1(sin x), and negative-argument and reciprocal relations.

Figure 1
The trigonometric functions are periodic, so they are many-one and have no inverse on their whole domains. To invert them, restrict each to an interval on which it is one-one and still takes every value in its range. The chosen interval becomes the range of the inverse function, and its values are called principal values. Principal value branches: sin⁡−1x\sin^{-1}x: domain [−1,1][-1, 1], range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. cos⁡−1x\cos^{-1}x: domain [−1,1][-1, 1], range [0,π][0, \pi]. tan⁡−1x\tan^{-1}x: domain R\mathbb{R}, range (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). cot⁡−1x\cot^{-1}x: domain R\mathbb{R}, range (0,π)(0, \pi). sec⁡−1x\sec^{-1}x: domain (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty), range [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}. csc⁡−1x\csc^{-1}x: domain (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty), range [−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}. The graphs of sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1} are shown in the figure; each is the reflection in y=xy = x of the restricted branch of the original function. sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1} are increasing; cos⁡−1\cos^{-1} and cot⁡−1\cot^{-1} are decreasing. tan⁡−1x\tan^{-1}x has horizontal asymptotes y=±π2y = \pm\frac{\pi}{2}. Note that sin⁡−1x\sin^{-1}x means the inverse function, not 1sin⁡x\frac{1}{\sin x}, which is csc⁡x\csc x. Composition in the two orders. sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x for every x∈[−1,1]x \in [-1, 1]: applying sin⁡\sin undoes the inverse. sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x only for x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]. For other xx, the answer is the unique angle in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] with the same sine. For example, sin⁡−1(sin⁡5π6)=π6\sin^{-1}(\sin\frac{5\pi}{6}) = \frac{\pi}{6}, and sin⁡−1(sin⁡2)=π−2\sin^{-1}(\sin 2) = \pi - 2. The graph of y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) is a zigzag (triangular wave) with slopes ±1\pm 1 and period 2π2\pi. Similarly cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x only on [0,π][0, \pi], with cos⁡−1(cos⁡x)=2π−x\cos^{-1}(\cos x) = 2\pi - x on [π,2π][\pi, 2\pi]; and tan⁡−1(tan⁡x)=x−nπ\tan^{-1}(\tan x) = x - n\pi when x∈(nπ−π2,nπ+π2)x \in (n\pi - \frac{\pi}{2}, n\pi + \frac{\pi}{2}). Negative arguments. sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x, tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x, csc⁡−1(−x)=−csc⁡−1x\csc^{-1}(-x) = -\csc^{-1}x (odd functions). cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x, cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x) = \pi - \cot^{-1}x, sec⁡−1(−x)=π−sec⁡−1x\sec^{-1}(-x) = \pi - \sec^{-1}x. The difference comes from the ranges: symmetric about 00 for the first group, about π2\frac{\pi}{2} for the second. Reciprocal relations. csc⁡−1x=sin⁡−11x\csc^{-1}x = \sin^{-1}\frac{1}{x} and sec⁡−1x=cos⁡−11x\sec^{-1}x = \cos^{-1}\frac{1}{x} for ∣x∣≥1|x| \ge 1. cot⁡−1x=tan⁡−11x\cot^{-1}x = \tan^{-1}\frac{1}{x} for x>0x > 0, but cot⁡−1x=π+tan⁡−11x\cot^{-1}x = \pi + \tan^{-1}\frac{1}{x} for x<0x < 0 (because the ranges differ). Converting between inverse functions. Draw a right triangle (for positive arguments): if θ=sin⁡−1x\theta = \sin^{-1}x with 0<x<10 < x < 1, the opposite side is xx and the hypotenuse 11, so cos⁡θ=1−x2\cos\theta = \sqrt{1 - x^{2}} and tan⁡θ=x1−x2\tan\theta = \frac{x}{\sqrt{1 - x^{2}}}. Hence sin⁡−1x=tan⁡−1x1−x2\sin^{-1}x = \tan^{-1}\frac{x}{\sqrt{1 - x^{2}}} for ∣x∣<1|x| < 1. Key Takeaways: - Ranges: sin⁡−1,tan⁡−1\sin^{-1}, \tan^{-1} in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]; cos⁡−1,cot⁡−1\cos^{-1}, \cot^{-1} in [0,π][0, \pi]. - sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x always; sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x only on [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. - sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x; cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x. - cot⁡−1x=tan⁡−11x\cot^{-1}x = \tan^{-1}\frac{1}{x} only for x>0x > 0. - sin⁡−1x≠1sin⁡x\sin^{-1}x \neq \frac{1}{\sin x}. - Use a right triangle to convert one inverse function into another. - Pitfall: answering sin⁡−1(sin⁡3π4)=3π4\sin^{-1}(\sin\frac{3\pi}{4}) = \frac{3\pi}{4} instead of π4\frac{\pi}{4}. Brain-Teaser 1: Evaluate sin⁡−1(sin⁡10)\sin^{-1}(\sin 10) and cos⁡−1(cos⁡10)\cos^{-1}(\cos 10), with 1010 in radians. Solution: 10≈3.18π10 \approx 3.18\pi. For sine, find the angle in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] with the same sine: sin⁡10=sin⁡(3π−10)\sin 10 = \sin(3\pi - 10), and 3π−10≈−0.5753\pi - 10 \approx -0.575, which lies in the range, so sin⁡−1(sin⁡10)=3π−10\sin^{-1}(\sin 10) = 3\pi - 10. For cosine, cos⁡10=cos⁡(10−4π)=cos⁡(4π−10)\cos 10 = \cos(10 - 4\pi) = \cos(4\pi - 10), and 4π−10≈2.566∈[0,π]4\pi - 10 \approx 2.566 \in [0, \pi], so cos⁡−1(cos⁡10)=4π−10\cos^{-1}(\cos 10) = 4\pi - 10. Locating the angle in the principal range is the entire method. Brain-Teaser 2: Find cos⁡(tan⁡−134)\cos(\tan^{-1}\frac{3}{4}) and sin⁡(cot⁡−1(−34))\sin(\cot^{-1}(-\frac{3}{4})). Solution: For θ=tan⁡−134\theta = \tan^{-1}\frac{3}{4} (first quadrant), a 33-44-55 triangle gives cos⁡θ=45\cos\theta = \frac{4}{5}. For ϕ=cot⁡−1(−34)\phi = \cot^{-1}(-\frac{3}{4}), the range (0,π)(0, \pi) puts ϕ\phi in the second quadrant with cot⁡ϕ=−34\cot\phi = -\frac{3}{4}, so sin⁡ϕ=45\sin\phi = \frac{4}{5} (positive in the second quadrant). The sign comes from the principal range, not from the triangle. Brain-Teaser 3: Sketch y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) on [0,2π][0, 2\pi] and find its value at x=7π6x = \frac{7\pi}{6}. Solution: On [0,π2][0, \frac{\pi}{2}], y=xy = x; on [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}], y=π−xy = \pi - x (a line of slope −1-1 from π2\frac{\pi}{2} down to −π2-\frac{\pi}{2}); on [3π2,2π][\frac{3\pi}{2}, 2\pi], y=x−2πy = x - 2\pi. At x=7π6x = \frac{7\pi}{6}, y=π−7π6=−π6y = \pi - \frac{7\pi}{6} = -\frac{\pi}{6}. Check: sin⁡7π6=−12\sin\frac{7\pi}{6} = -\frac{1}{2} and sin⁡−1(−12)=−π6\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}. The zigzag graph makes such evaluations immediate.

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