Values of inverse trig expressions

A function and its inverse cancel, like 3 × 3⁻¹ = 1. So sin(sin⁻¹x) = x for every x in [–1, 1].

The other way round needs care: sin⁻¹(sin θ) = θ only when θ is already in the branch [–π/2, π/2]. The answer of sin⁻¹ can never fall outside its branch.

• sin⁻¹(sin π/5) = π/5, since π/5 is in [–π/2, π/2].
• sin⁻¹(sin 4π/5) is not 4π/5. First rewrite the angle: sin(4π/5) = sin(π – 4π/5) = sin(π/5). Now π/5 is in the branch, so the answer is π/5.

So the method is always: move the angle into the branch first, then cancel. The tools are the usual angle facts:

• sine: sin(π – θ) = sin θ;
• cosine: cos(2π – θ) = cos θ and cos(–θ) = cos θ;
• tangent: tan repeats every π, so tan(θ – π) = tan θ.

Example: cos⁻¹(cos 5π/3). 5π/3 is outside [0, π]. cos(5π/3) = cos(2π – 5π/3) = cos(π/3), and π/3 is inside, so the answer is π/3.

Example: tan⁻¹(tan 5π/4). tan repeats every π, so tan(5π/4) = tan(π/4), and π/4 is in (–π/2, π/2). The answer is π/4.

For sums of inverse values, find each principal value separately and then add. tan⁻¹(√3) + cos⁻¹(0) = π/3 + π/2 = 5π/6.

For nested expressions, work from the inside out. cos(2 sin⁻¹(½)): sin⁻¹(½) = π/6, so this is cos(π/3) = ½.