Trigonometric functions like sin, cos, tan are not one-one over their natural domains, so their inverses do not exist globally. By restricting their domains, we make them bijective and define their inverse functions. These inverse functions are critical for solving equations and appear frequently in calculus.
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Key Concepts
Principal Value Branches (Domains and Ranges)
| Function | Domain | Range (Principal Branch) |
|---|---|---|
| sin-1 x | [-1, 1] | [-pi/2, pi/2] |
| cos-1 x | [-1, 1] | [0, pi] |
| tan-1 x | R | (-pi/2, pi/2) |
| cosec-1 x | R - (-1,1) | [-pi/2, pi/2] - {0} |
| sec-1 x | R - (-1,1) | [0, pi] - {pi/2} |
| cot-1 x | R | (0, pi) |
- Important Properties:
- sin(sin-1 x) = x for x in [-1, 1]
- sin-1(sin x) = x for x in [-pi/2, pi/2]
- sin-1 x + cos-1 x = pi/2 for x in [-1, 1]
- tan-1 x + cot-1 x = pi/2 for all x in R
- sec-1 x + cosec-1 x = pi/2 for |x| ≥ 1
- Addition Formulae:
- tan-1 x + tan-1 y = tan-1((x+y)/(1-xy)) when xy < 1
- tan-1 x - tan-1 y = tan-1((x-y)/(1+xy)) when xy > -1
- 2 tan-1 x = sin-1(2x/(1+x2)) = cos-1((1-x2)/(1+x2)) for |x| ≤ 1
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Worked Examples
Find the principal value of sin-1(1/2).
We need angle theta in [-pi/2, pi/2] such that sin(theta) = 1/2. Since sin(pi/6) = 1/2 and pi/6 is in the principal branch, sin-1(1/2) = pi/6.
Evaluate cos-1(-1/2).
We need angle in [0, pi] such that cos(theta) = -1/2. cos(2pi/3) = -1/2 and 2pi/3 is in [0, pi]. So cos-1(-1/2) = 2pi/3.
Evaluate tan-1(1) + sin-1(-1/2).
tan-1(1) = pi/4 (since tan(pi/4) = 1).
sin-1(-1/2) = -pi/6 (since sin(-pi/6) = -1/2 and -pi/6 is in [-pi/2, pi/2]).
Sum = pi/4 + (-pi/6) = 3pi/12 - 2pi/12 = pi/12.
Simplify: sin-1(sin(7pi/6)).
Note that 7pi/6 is NOT in the principal branch [-pi/2, pi/2].
sin(7pi/6) = -1/2. Now sin-1(-1/2) = -pi/6.
So sin-1(sin(7pi/6)) = -pi/6.
Prove that tan-1(1/2) + tan-1(1/3) = pi/4.
Using addition formula with x = 1/2, y = 1/3: xy = 1/6 < 1.
tan-1((1/2 + 1/3)/(1 - 1/6)) = tan-1((5/6)/(5/6)) = tan-1(1) = pi/4. Proved.
Evaluate cos(tan-1(3/4)).
Let theta = tan-1(3/4), so tan(theta) = 3/4.
Draw a right triangle: opposite = 3, adjacent = 4, hypotenuse = 5.
cos(theta) = 4/5.
Write the simplified form of tan-1(cos x/(1 - sin x)).
Use identities: cos x = cos2(x/2) - sin2(x/2), 1 - sin x = (cos(x/2) - sin(x/2))2.
Expression becomes tan-1((cos(x/2) + sin(x/2))/(cos(x/2) - sin(x/2))) = tan-1(tan(pi/4 + x/2)) = pi/4 + x/2.
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Common mistakes
- Forgetting to check whether the angle lies in the principal value branch before applying sin-1(sin x) = x.
- Assuming sin-1 x = 1/sin x. These are very different — sin-1 x is the inverse function, not the reciprocal.
- Using the addition formula for tan-1 without verifying the condition xy < 1 or xy > -1.
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Summary
Inverse trig functions are defined on restricted domains to ensure they are bijective. Their principal value branches and key identities are essential for simplification and problem-solving in higher mathematics.