Trigonometric Identities

Michigan State University - Diagnostic Exam

Let $x = sinθ$ with $0 ≤ θ ≤ π- 2$ . Which of the following is equivalent to the expression $sin2-θ+ θ 2$ ?

a) arcsin x + x b) arcsin x + x c) + x

d) + x e) + x

Solution:

Notice that $cos θ ≥ 0$ and $√ ------- cos θ = 1 - x2$ so that

sin-2θ 2sinθ-cos-θ 2 + θ = 2 + arcsin x √ -----2- = x 1 - x + arcsin x

Diagnostic:

If you miss this question please review inverse trigonometric functions and double angle formulas.

July 15, 2008