Principal value branches

From Chapter 1: only a bijection (one-one and onto) has an inverse. The trigonometric functions are not one-one, because they repeat: sin 30°, sin 150° and sin 390° are all ½. So 'the angle whose sine is ½' has endless answers.

The fix is to restrict each function to one piece of its graph where it rises (or falls) steadily, taking every value exactly once. On that piece it is a bijection, and it can be inverted. The chosen piece is called the principal value branch.

For sin x, the piece from –π/2 to π/2 works: sine climbs from –1 to 1 there, never repeating. So sin⁻¹x takes inputs from [–1, 1] and gives answers in [–π/2, π/2].

The same idea gives all six branches. The range is where the answer must lie:

You do not need to memorise this as a list. Look at each graph: the end points are closed brackets where the function has a value, and open where it shoots off to infinity (tan at ±π/2, cot at 0 and π). Cosec and sec leave out the one angle where they are undefined (0 for cosec, π/2 for sec).

The principal value of sin⁻¹x is the one angle in the branch whose sine is x. Two steps: find an angle with the right value, then move it into the branch.

Example: sin⁻¹(–√3/2). We know sin(π/3) = √3/2. The branch [–π/2, π/2] allows negative angles, and sine is odd, so sin(–π/3) = –√3/2. The answer is –π/3, not 4π/3.

Example: cos⁻¹(–√3/2). cos(π/6) = √3/2, and the branch [0, π] has no negative angles, so use cos(π – θ) = –cos θ: cos(5π/6) = –√3/2. The answer is 5π/6.

A warning about notation: sin⁻¹x means 'the angle whose sine is x'. It is not 1/sin x, which is written (sin x)⁻¹ or cosec x.