Undoing an inverse

From the definition: sin(sin⁻¹ x) = x for every x ∈ [−1, 1], and sin⁻¹(sin x) = x for x ∈ [−π/2, π/2]. The same pairs hold for the other five functions on their own domains and principal ranges.

The first form is always safe inside the domain: cos(cos⁻¹ 0.3) = 0.3 and tan(tan⁻¹ 5) = 5.

The second form fails outside the principal range, because the inverse can only return a principal angle. sin⁻¹(sin 2π/3) is π/3, not 2π/3.

Method: replace the angle by one in the principal range with the same trigonometric value. sin 3π/5 = sin(π − 3π/5) = sin 2π/5, and 2π/5 lies in [−π/2, π/2], so sin⁻¹(sin 3π/5) = 2π/5.

tan⁻¹(tan 3π/4): tan 3π/4 = −1, and tan⁻¹(−1) = −π/4. cos⁻¹(cos 7π/6): cos 7π/6 = −√3/2 and cos⁻¹(−√3/2) = 5π/6.

Drawn as a function of x over all of R, y = sin⁻¹(sin x) is a zigzag: it follows y = x on [−π/2, π/2], then y = π − x on [π/2, 3π/2], and so on, never leaving [−π/2, π/2].

Mixed forms reduce to a right triangle. If θ = tan⁻¹ x with |x| < 1, the triangle has opposite x, adjacent 1 and hypotenuse √(1 + x²), so sin(tan⁻¹ x) = x/√(1 + x²).