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MHT-CET Maths · Trigonometric Functions

Inverse Trigonometric Functions — Principal Values and Evaluating Expressions

Each inverse trigonometric function returns one angle from a fixed principal range, so evaluating an expression means placing every inverse value in its own range and then converting between ratios with a right triangle.

All Trigonometric Functions notesMHT-CET Maths strategy

Why this matters

32 PYQs, and the page where inverse trigonometry is won or lost: 15 test the principal ranges directly (a value, a sum of values, a domain, an inequality), and 17 ask for a trigonometric ratio of an inverse value or of a sum of two. Nothing here is long; the marks go to the student who knows that sin⁻¹(sin 2π/3) is not 2π/3.

Concept 1 of 2: Principal Ranges, Negative Arguments and f⁻¹(f(x))

An inverse function must give ONE answer, so each is restricted to a range: sine and tangent to angles around 0, cosine and cotangent to [0,π][0, \pi]. Every principal-value question is a test of these ranges — which is why sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta only when θ\theta is already inside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Definition

  • 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; but 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.
  • f−1(f(θ))f^{-1}(f(\theta)): find the angle IN the range with the same ratio. sin⁡−1(sin⁡2π3)=π3\sin^{-1}(\sin\frac{2\pi}{3}) = \frac{\pi}{3}; tan⁡−1(tan⁡7π6)=π6\tan^{-1}(\tan\frac{7\pi}{6}) = \frac{\pi}{6}; cos⁡−1(cos⁡23π20)=17π20\cos^{-1}(\cos\frac{23\pi}{20}) = \frac{17\pi}{20}.
  • Extremes: cos⁡−1x≤π\cos^{-1}x \le \pi, so cos⁡−1x+cos⁡−1y+cos⁡−1z=3π\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = 3\pi forces each to be π\pi, i.e. x=y=z=−1x = y = z = -1.
  • Domains: sin⁡−1u\sin^{-1}u, cos⁡−1u\cos^{-1}u need −1≤u≤1-1 \le u \le 1. For sin⁡−1(2x)+π6\sqrt{\sin^{-1}(2x) + \frac{\pi}{6}} also need sin⁡−1(2x)≥−π6\sin^{-1}(2x) \ge -\frac{\pi}{6}, so −14≤x≤12-\frac14 \le x \le \frac12.
  • Approximating an inverse value (tan⁡−10.999\tan^{-1}0.999) is a differentials question, taught in Applications of Derivative: tan⁡−1(1+h)≈π4+h2\tan^{-1}(1 + h) \approx \frac{\pi}{4} + \frac{h}{2}.
FunctionDomainPrincipal range
sin⁡−1x\sin^{-1}x[−1,1][-1, 1][−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Q
cos⁡−1x\cos^{-1}x[−1,1][-1, 1][0,π][0, \pi]Q
tan⁡−1x\tan^{-1}xR\mathbb{R}(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)Q
cot⁡−1x\cot^{-1}xR\mathbb{R}(0,π)(0, \pi)Q
sec⁡−1x\sec^{-1}x∣x∣≥1|x| \ge 1[0,π][0, \pi], not π2\frac{\pi}{2}Q
csc⁡−1x\csc^{-1}x∣x∣≥1|x| \ge 1[−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], not 0Q
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q132Moderate

Example 1 · Trigonometric Functions · Inverse Trigonometric Functions — Principal Values and Evaluation

The value of tan⁡−1(1)+cos⁡−1 ⁣(−12)+sin⁡−1 ⁣(−12)\tan^{-1}(1)+\cos^{-1}\!\left(-\dfrac{1}{2}\right)+\sin^{-1}\!\left(-\dfrac{1}{2}\right) is

Taking the negative out of cos⁻¹

cos⁡−1(−12)\cos^{-1}\left(-\frac12\right) is 2π3\frac{2\pi}{3}, not −π3-\frac{\pi}{3}. Negative angles are outside the range of cos⁡−1\cos^{-1}, cot⁡−1\cot^{-1} and sec⁡−1\sec^{-1}; for these three the rule is π\pi minus the positive value.

Several options can be true

With α=3sin⁡−1611≈1.74\alpha = 3\sin^{-1}\frac{6}{11}\approx 1.74 and β=3cos⁡−149≈3.33\beta = 3\cos^{-1}\frac49 \approx 3.33, the statements sin⁡β<0\sin\beta < 0, cos⁡(α+β)>0\cos(\alpha + \beta) > 0 and cos⁡α<0\cos\alpha < 0 are all true. The paper keys sin⁡β<0\sin\beta < 0; place each angle by estimate before choosing.
Drill 14 more on principal ranges, negative arguments and f⁻¹(f(x))

Concept 2 of 2: A Trigonometric Ratio of an Inverse Value — the Right-Triangle Conversion

sin⁡−135\sin^{-1}\frac35 is just an angle whose sine is 35\frac35. Draw the right triangle with opposite 3, hypotenuse 5 and adjacent 4, and every other ratio of that angle is read off. Sums of two inverse values then fall to the compound-angle formula.

Definition

  • Convert by a right triangle: sin⁡−135=cos⁡−145=tan⁡−134\sin^{-1}\frac35 = \cos^{-1}\frac45 = \tan^{-1}\frac34 (for positive arguments).
  • Compositions in xx: sin⁡(cot⁡−1x)=11+x2\sin(\cot^{-1}x) = \dfrac{1}{\sqrt{1 + x^2}}, cos⁡(tan⁡−1x)=11+x2\cos(\tan^{-1}x) = \dfrac{1}{\sqrt{1 + x^2}}, sec⁡2(tan⁡−1x)=1+x2\sec^2(\tan^{-1}x) = 1 + x^2, csc⁡2(cot⁡−1x)=1+x2\csc^2(\cot^{-1}x) = 1 + x^2.
  • Sums: cos⁡(sin⁡−1a+cos⁡−1b)\cos(\sin^{-1}a + \cos^{-1}b) — let α=sin⁡−1a\alpha = \sin^{-1}a, β=cos⁡−1b\beta = \cos^{-1}b, read all four ratios from two triangles, then expand cos⁡(α+β)\cos(\alpha + \beta).
  • Doubles: sin⁡(2sin⁡−1x)=2x1−x2\sin(2\sin^{-1}x) = 2x\sqrt{1 - x^2}; tan⁡(2tan⁡−1x)=2x1−x2\tan(2\tan^{-1}x) = \frac{2x}{1 - x^2}.
  • Negative arguments change the sign of one leg: cos⁡−1(−35)\cos^{-1}\left(-\frac35\right) has cosine −35-\frac35 and sine +45+\frac45, so sin⁡(2cos⁡−1(−35))=−2425\sin\left(2\cos^{-1}\left(-\frac35\right)\right) = -\frac{24}{25}.

Two conversions worth memorising

sin⁡(cot⁡−1x)=cos⁡(tan⁡−1x)=11+x2sin⁡(2sin⁡−1x)=2x1−x2\sin(\cot^{-1}x)=\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}} \qquad \sin(2\sin^{-1}x)=2x\sqrt{1-x^2}

Worked example

Evaluate tan⁡(sin⁡−145+tan⁡−113)\tan\left(\sin^{-1}\frac45 + \tan^{-1}\frac13\right).
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q101Moderate

Example 2 · Trigonometric Functions · Inverse Trigonometric Functions — Principal Values and Evaluation

The value of tan⁡ ⁣(sin⁡−1 ⁣35+tan⁡−1 ⁣23)\tan\!\left(\sin^{-1}\!\frac{3}{5}+\tan^{-1}\!\frac{2}{3}\right) is

Doubling the ratio instead of the angle

sin⁡(2sin⁡−10.8)\sin(2\sin^{-1}0.8) is 2(0.8)(0.6)=0.962(0.8)(0.6) = 0.96, not 2×0.82 \times 0.8. The double-angle formula needs the cosine too, and the option 0.160.16 or 1.61.6 catches the shortcut.
Drill 16 more on a trigonometric ratio of an inverse value — the right-triangle conversion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Reference tables (1)

Principal Ranges, Negative Arguments and f⁻¹(f(x))6 rows
FunctionDomainPrincipal range
sin⁡−1x\sin^{-1}x[−1,1][-1, 1][−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Q
cos⁡−1x\cos^{-1}x[−1,1][-1, 1][0,π][0, \pi]Q
tan⁡−1x\tan^{-1}xR\mathbb{R}(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)Q
cot⁡−1x\cot^{-1}xR\mathbb{R}(0,π)(0, \pi)Q
sec⁡−1x\sec^{-1}x∣x∣≥1|x| \ge 1[0,π][0, \pi], not π2\frac{\pi}{2}Q
csc⁡−1x\csc^{-1}x∣x∣≥1|x| \ge 1[−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], not 0Q

Watch out for (3)

Test yourself on Trigonometric Functions

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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