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๐Ÿ”บTrigonometry Unit 8 Review

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8.3 Trigonometric Equations with Inverse Functions

8.3 Trigonometric Equations with Inverse Functions

Written by the Fiveable Content Team โ€ข Last updated August 2025
Written by the Fiveable Content Team โ€ข Last updated August 2025
๐Ÿ”บTrigonometry
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Inverse trigonometric functions are powerful tools for solving equations. They undo trig operations, letting us isolate variables and find solutions. But we must be careful about domains and ranges to get accurate answers.

Using inverse trig functions requires strategy. We isolate trig terms, apply the inverse, and consider multiple solutions. Properties and identities help simplify complex equations. Always verify your answers by plugging them back in.

Solving Trigonometric Equations with Inverse Functions

Inverse trigonometric equation solutions

  • Inverse trigonometric functions undo trigonometric operations (sinโกโˆ’1x\sin^{-1}x, cosโกโˆ’1x\cos^{-1}x, tanโกโˆ’1x\tan^{-1}x, cscโกโˆ’1x\csc^{-1}x, secโกโˆ’1x\sec^{-1}x, cotโกโˆ’1x\cot^{-1}x)
  • Apply inverse functions when equation isolates trig function (solve sinโกx=0.5\sin x = 0.5 with x=sinโกโˆ’1(0.5)x = \sin^{-1}(0.5))
  • Use composition properties simplify equations (sinโก(sinโกโˆ’1x)=x\sin(\sin^{-1}x) = x for xx in [โˆ’1,1][-1,1], sinโกโˆ’1(sinโกx)=x\sin^{-1}(\sin x) = x for xx in [โˆ’ฯ€2,ฯ€2][-\frac{\pi}{2}, \frac{\pi}{2}])
  • Isolate trig function before applying inverse (solve 2sinโกx+1=02\sin x + 1 = 0 as sinโกx=โˆ’0.5\sin x = -0.5, then x=sinโกโˆ’1(โˆ’0.5)x = \sin^{-1}(-0.5))
  • Consider multiple solutions within domain (equation sinโกx=0.5\sin x = 0.5 has solutions x=ฯ€6x = \frac{\pi}{6} and x=5ฯ€6x = \frac{5\pi}{6})
Inverse trigonometric equation solutions, Inverse Trigonometric Functions โ€น OpenCurriculum

Properties of inverse trig functions

  • Inverse function relationships simplify equations (tanโกโˆ’1x+cotโกโˆ’1x=ฯ€2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, sinโกโˆ’1x+cosโกโˆ’1x=ฯ€2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2})
  • Apply identities with inverse functions (tanโกโˆ’1x+tanโกโˆ’1y=tanโกโˆ’1(x+y1โˆ’xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\frac{x+y}{1-xy}) for xy<1xy < 1)
  • Odd and even properties aid simplification (sinโกโˆ’1(โˆ’x)=โˆ’sinโกโˆ’1x\sin^{-1}(-x) = -\sin^{-1}x, cosโกโˆ’1(โˆ’x)=ฯ€โˆ’cosโกโˆ’1x\cos^{-1}(-x) = \pi - \cos^{-1}x)
  • Simplify expressions using properties (rewrite sinโกโˆ’1(sinโกx)\sin^{-1}(\sin x) as xx for xx in [โˆ’ฯ€2,ฯ€2][-\frac{\pi}{2}, \frac{\pi}{2}])
Inverse trigonometric equation solutions, Inverse Trigonometric Functions | Precalculus

Advanced Techniques and Considerations

Domain and range in inverse trig

  • Restricted domains ensure unique inverse function values:
    1. sinโกโˆ’1x:[โˆ’1,1]โ†’[โˆ’ฯ€2,ฯ€2]\sin^{-1}x: [-1, 1] \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}]
    2. cosโกโˆ’1x:[โˆ’1,1]โ†’[0,ฯ€]\cos^{-1}x: [-1, 1] \rightarrow [0, \pi]
    3. tanโกโˆ’1x:(โˆ’โˆž,โˆž)โ†’(โˆ’ฯ€2,ฯ€2)\tan^{-1}x: (-\infty, \infty) \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2})
  • Domain restrictions affect solution sets (equation cosโกโˆ’1x=2ฯ€\cos^{-1}x = 2\pi has no solution as range is [0,ฯ€][0, \pi])
  • Solutions outside principal range need adjustment (sinโกโˆ’1(0.5)=ฯ€6\sin^{-1}(0.5) = \frac{\pi}{6} or 5ฯ€6\frac{5\pi}{6}, but 5ฯ€6\frac{5\pi}{6} not in principal range)
  • Fit solutions within appropriate domain (adjust tanโกโˆ’1(1)+2ฯ€=ฯ€4+2ฯ€\tan^{-1}(1) + 2\pi = \frac{\pi}{4} + 2\pi to ฯ€4\frac{\pi}{4} in (โˆ’ฯ€2,ฯ€2)(-\frac{\pi}{2}, \frac{\pi}{2}))

Complex equations with inverse trig

  • Substitution simplifies equations (let u=sinโกโˆ’1xu = \sin^{-1}x in sinโก(sinโกโˆ’1x)+cosโก(sinโกโˆ’1x)=1\sin(\sin^{-1}x) + \cos(\sin^{-1}x) = 1)
  • Isolate trig terms algebraically (solve tanโกโˆ’1x+tanโกโˆ’1y=ฯ€4\tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} for yy in terms of xx)
  • Apply inverse functions strategically (solve sinโก2x=cosโกx\sin 2x = \cos x as x=12sinโกโˆ’1(cosโกx)x = \frac{1}{2}\sin^{-1}(\cos x))
  • Combine identities with inverse properties (use sinโก2x+cosโก2x=1\sin^2 x + \cos^2 x = 1 with sinโกโˆ’1\sin^{-1} and cosโกโˆ’1\cos^{-1})
  • Verify solutions by substitution (check if x=ฯ€6x = \frac{\pi}{6} satisfies sinโก2x=cosโกx\sin 2x = \cos x)
  • Handle multiple inverse functions (solve sinโกโˆ’1x+cosโกโˆ’1x=ฯ€2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2})
  • Solve composite inverse function equations (sinโกโˆ’1(cosโก(tanโกโˆ’1x))=ฯ€3\sin^{-1}(\cos(\tan^{-1}x)) = \frac{\pi}{3})