Advanced Trigonometry

This lesson continues from Trigonometry — Right Triangles, Unit Circle, and Graphs. You should already be comfortable with radians, exact unit-circle values, graph parameters, and the Pythagorean identity.

Reciprocal and Quotient Identities

sec θ = 1/cos θ      csc θ = 1/sin θ      cot θ = 1/tan θ
tan θ = sin θ/cos θ                     cot θ = cos θ/sin θ
sin² θ + cos² θ = 1
1 + tan² θ = sec² θ
1 + cot² θ = csc² θ

An identity is established by transforming one side into the other. Do not perform unrelated operations on both sides as though solving an equation.

Worked example: verify an identity

Verify (1 − cos2 x)/sin x = sin x wherever both sides are defined.

(1 − cos² x)/sin x
= sin² x/sin x       because 1 − cos² x = sin² x
= sin x              provided sin x ≠ 0

Compound-Angle and Double-Angle Formulas

sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B

sin 2A = 2 sin A cos A
cos 2A = cos² A − sin² A
        = 2cos² A − 1
        = 1 − 2sin² A

Worked example: exact value

Find sin 75° exactly by writing 75° = 45° + 30°.

sin 75° = sin(45° + 30°)
        = sin 45° cos 30° + cos 45° sin 30°
        = (√2/2)(√3/2) + (√2/2)(1/2)
        = (√6 + √2)/4

Solving Trigonometric Equations

Solve for a basic trig value, find the reference angle, choose every quadrant with the correct sign, and include all solutions in the requested interval.

Worked example: solutions over one cycle

Solve 2 sin2 x − sin x − 1 = 0 for 0 ≤ x < 2π.

Let u = sin x.
2u² − u − 1 = 0
(2u + 1)(u − 1) = 0

sin x = −1/2 or sin x = 1

sin x = −1/2 at x = 7π/6, 11π/6
sin x = 1 at x = π/2

Solutions: π/2, 7π/6, 11π/6

Law of Sines and Law of Cosines

a/sin A = b/sin B = c/sin C
c² = a² + b² − 2ab cos C

Use the Law of Cosines for SAS or SSS information. Use the Law of Sines when you know an angle and its opposite side. With SSA data, a second triangle may exist.

Worked example: a non-right triangle

Two sides are 7 cm and 10 cm with included angle 60°. Find the opposite side c.

c² = 7² + 10² − 2(7)(10)cos 60°
   = 49 + 100 − 140(1/2)
   = 79
c = √79
c ≈ 8.89 cm

The result lies between the difference 3 and sum 17 of the other sides, so it passes the triangle-inequality check.

Practice Set

  1. Simplify sec x cos x where defined.
  2. Verify (sec2 x − 1)/tan x = tan x where defined.
  3. Find cos 15° exactly using 45° − 30°.
  4. Solve 2 cos x + √3 = 0 for 0 ≤ x < 2π.
  5. Two sides of a triangle are 5 m and 8 m, with included angle 120°. Find the third side.

Answer Checks

  1. sec x cos x = (1/cos x)cos x = 1, for cos x ≠ 0.
  2. sec2 x − 1 = tan2 x, so the quotient is tan x where tan x is defined and nonzero.
  3. cos 15° = cos(45° − 30°) = (√6 + √2)/4.
  4. cos x = −√3/2, so x = 5π/6 or 7π/6.
  5. c2 = 52 + 82 − 2(5)(8)cos 120° = 129, so c = √129 ≈ 11.36 m.