Inverse trigonometric functions

Maths · Inverse trigonometric functions. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

sin⁻¹x means the angle whose sine is x. It does not mean 1/sin x. Many angles have the same sine, so each trigonometric function is first restricted to a part of its domain where it is one-to-one. The inverse then gives exactly one angle, called the principal value.

You must know the domain and range of each inverse function, because the answer has to lie in the principal range. A negative x gives a negative angle for sin⁻¹ and tan⁻¹, but an angle in the second quadrant for cos⁻¹.

For expressions such as sin(cos⁻¹x), call the inner part θ and work with a right-angled triangle or an identity.

Rules to remember

y = sin⁻¹x: domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2
y = cos⁻¹x: domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π
y = tan⁻¹x: domain all real numbers, range −π/2 < y < π/2
sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹A + tan⁻¹B = tan⁻¹((A + B)/(1 − AB)) and tan⁻¹A − tan⁻¹B = tan⁻¹((A − B)/(1 + AB))
sin⁻¹A + sin⁻¹B = sin⁻¹(A√(1 − B²) + B√(1 − A²)) and cos⁻¹A + cos⁻¹B = cos⁻¹(AB − √((1 − A²)(1 − B²)))
sin⁻¹x + cos⁻¹x = π/2 and 2 tan⁻¹A = tan⁻¹(2A/(1 − A²))

Common mistake

Giving an angle outside the principal range, for example writing sin⁻¹(−1/2) as 7π/6 or 11π/6. The correct value is −π/6, because sin⁻¹ must lie between −π/2 and π/2.

Worked example

Find the value of tan(cos⁻¹(−5/13)).

Let θ = cos⁻¹(−5/13). Then cos θ = −5/13 and θ lies between π/2 and π (second quadrant).
sin²θ = 1 − 25/169 = 144/169, and sine is positive in the second quadrant, so sin θ = 12/13.
tan θ = sin θ ÷ cos θ = (12/13) ÷ (−5/13) = −12/5.

Answer: tan(cos⁻¹(−5/13)) = −12/5.

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