Concepts: identities, simplification and equations: Inverse Trigonometric Functions

3.1 Complementary identities carry easy marks

The most used results in this chapter are the three complementary pairs. Together with the negative angle results they handle a large share of board questions in one line.

For sin inverse x plus cos inverse x, since their ranges sit inside the first quadrant pieces, the sum is always pi/2 for x in [-1, 1]. The tangent and cotangent pair works for every real x, and the secant and cosecant pair works outside the open interval from -1 to 1. When a question asks for sin inverse of a number plus cos inverse of the same number, write pi/2 immediately, but say why in one line.

Negative arguments need care. Pulling a minus sign out is legal for sine, tangent and cosecant inverses. For cosine, cotangent and secant inverses the result changes to pi minus the original value, so cos inverse(-1/2) is pi - cos inverse(1/2) which is pi - pi/3 = 2pi/3.

3.2 Simplifying expressions with a triangle

When an inverse function appears inside another inverse, or inside a trigonometric function, draw a right triangle. For tan inverse x, mark the opposite side x and the adjacent side 1. The hypotenuse becomes square root of (1 + x^2), and every ratio can be read off.

This turns tan inverse of the ratio cos x over (1 - sin x) into a simple angle. Multiply the top and bottom by (1 + sin x). The denominator becomes 1 - sin^2 x, which is cos^2 x. After cancelling the common cos x the expression is (1 + sin x)/cos x, that is sec x + tan x. Recognising sec x + tan x as tan(pi/4 + x/2) gives the simplest form pi/4 + x/2 for x inside the principal range of tangent.

The triangle method also proves the odd-looking identity tan inverse of square root x equals half of cos inverse of (1 - x)/(1 + x) for x in [0, 1]. Put the cosine value equal to a new angle and use the half angle formula for tangent; the square root of x appears on both sides.

3.3 Equations that mix inverse functions

Equations usually combine the tangent sum formula with the principal range. For tan inverse 2x plus tan inverse 3x equal to pi/4, take tangent of both sides. The left side becomes (2x + 3x)/(1 - 6x^2), so 5x/(1 - 6x^2) = 1, giving 6x^2 + 5x - 1 = 0 and then x = 1/6 or x = -1. The candidate x = -1 must be tested, because tan inverse(-2) + tan inverse(-3) is negative, while pi/4 is positive. So only x = 1/6 survives. Testing candidates against the principal range is where most students lose marks.

3.4 Exercise coverage

The property-based exercises of the chapter ask for principal values of sums, simplification using complementary results, and solving equations. The worked items on this page and the questions page represent Exercise 2.2 and the Miscellaneous Exercise, including the classic combinations with pi/2, 2pi/3 and pi/4.

3.5 Three worked simplifications

Work through the common shapes before attempting the exercise set. First, tan inverse(1/2) plus tan inverse(1/3). The product of the inputs is 1/6, below 1, so the sum formula applies and gives tan inverse of (1/2 + 1/3)/(1 - 1/6), that is tan inverse of 1, which is pi/4.

Second, sin inverse of sin of 4pi/5. The inner angle is outside [-pi/2, pi/2], but sin(4pi/5) equals sin(pi/5), and pi/5 is inside the branch, so the value is pi/5.

Third, cos inverse of cos of 7pi/6. The inner angle is outside [0, pi]. Since cos(7pi/6) equals cos(5pi/6), and 5pi/6 lies inside the branch, the value is 5pi/6. In each case the pattern is the same: match the value first, then choose the representative angle inside the branch.

3.5.1 Signs to check before simplifying

A simplification is incomplete until the sign is settled. For sin inverse of sin of an angle beyond the first quadrant, match the reference angle first and then decide whether the principal value is positive or negative. For cos inverse of a negative number the answer sits in the second quadrant, between pi by 2 and pi. For tan inverse of a negative number the answer is negative and lies between minus pi by 2 and 0. Board answer keys always show the interval or the sign, so write that line. Finally, when a question asks for the number of solutions inside a stated interval, draw the branch once and count the intersections instead of solving blindly.