Trigonometric Equations

Trigonometric Equations In quadratic form, using identities or linear in sine and cosine

Solving a Trig Equation in Quadratic Form • Solve the equation: • 2sin2θ – 3 sin θ + 1 = 0, 0 ≤ θ≤ 2p • Let sin θ equal some variable • sin θ = a • Factor this equation • (2a – 1) (a – 1) = 0 • Therefore a = ½ a = 1

Solving a Trig Equation in Quadratic Form • Now substitute sin θ back in for a • sin θ = ½ sin θ = 1 • Now do the inverse sin to find what θ equals • θ = sin-1 (½) θ = sin-1 1 • θ = p/6 and 5p/6 θ = p/2

Solving a Trig Equation in Quadratic Form • Solve the equation: • (tan θ – 1)(sec θ – 1) = 0 • tan θ – 1 = 0 sec θ – 1 = 0 • tan θ = 1 sec θ = 1 • θ = tan-1 1 θ = sec-1 1 • θ = p/4 and 5p/4 θ = 0

Solving a Trig Equation Using Identities • In order to solve trig equations, we want to have a single trig word in the equation. We can use trig identities to accomplish this goal. • Solve the equation • 3 cos θ + 3 = 2 sin2θ • Use the pythagorean identities to change sin2 θ to cos θ

Solving a Trig Equation Using Identities • sin2 = 1 – cos2 θ • Substituting into the equation • 3 cos θ + 3 = 2(1 – cos2θ) • To solve a quadratic equation it must be equal to 0 • 2cos2θ + 3 cos θ + 1 = 0 • Let cos θ = b

Solving a Trig Equation using Identities • 2b2 + 3b + 1 = 0 • (2b + 1) (b + 1) = 0 • (2b + 1) = 0 b + 1 = 0 • b = -½ b = -1 • cos θ = -½ cos θ = -1 • θ = 2p/3, 4p/3 θ = p

Solving a Trig Equation Using Identities • cos2θ – sin2θ + sin θ = 0 • 1 – sin2θ – sin2θ + sin θ = 0 • -2sin2 θ + sin θ + 1 = 0 • 2 sin2θ – sin θ – 1 = 0 • Let c = sin θ • 2c2 – c – 1 = 0 • (2c + 1) (c – 1) = 0

Solving a Trig Equation Using Identities • (2c + 1) = 0 c – 1 = 0 • c = -½ c = 1 • sin θ = -½ sin θ = 1 • θ = p/3 + p q=2p-p/3θ = p/2 • θ = 4p/3, q = 7p/3

Solving a Trig Equation Using Identities • Solve the equation • sin (2θ) sin θ = cos θ • Substitute in the formula for sin 2θ • (2sin θ cos θ)sin θ=cos θ • 2sin2 θ cos θ – cos θ = 0 • cos θ(2sin2 – 1) = 0 • cos θ = 0 2sin2 θ=1

Solving a Trig Equation Using Identities • cos θ = 0 • θ = 0, pθ = p/4, 3p/4, 5p/4, 7p/4

Solving a Trig Equation Using Identities • sin θ cos θ = -½ • This looks very much like the sin double angle formula. The only thing missing is the two in front of it. • So . . . multiply both sides by 2 • 2 sin θ cos θ = -1 • sin 2θ = -1 • 2 θ = sin-1 -1

Solving a Trig Equation Using Identities • 2θ = 3p/2 • θ = 3p/4 2θ = 3p/2 + 2p 2q = 7p/2 q = 7p/4

Solving a Trig Equation Linear in sin θ and cos θ • sin θ + cos θ = 1 • There is nothing I can substitute in for in this problem. The best way to solve this equation is to force a pythagorean identity by squaring both sides. • (sin θ + cos θ)2 = 12

Solving a Trig Equation Linear in sin θ and cos θ • sin2θ + 2sin θ cos θ + cos2 θ = 1 • 2sin θ cos θ + 1 = 1 • 2sin θ cos θ = 0 • sin 2θ = 0 • 2θ = 0 2θ = p • θ = 0 θ = p/2 • θ = p θ = 3p/2

Solving a Trig Equation Linear in sin θ and cos θ • Since we squared both sides, these answers may not all be correct (when you square a negative number it becomes positive). • In the original equation, there were no terms that were squared

Solving a Trig Equation Linear in sin θ and cos θ • Check: • Does sin 0 + cos 0 = 1? • Does sin p/2 + cos p/2 = 1? • Does sin p + cos p = 1? • Does sin 3p/2 + cos 3p/2 = 1?

Solving a Trig Equation Linear in sin θ and cos θ • sec θ = tan θ + cot θ • sec2 θ = (tan θ + cot θ)2 • sec2 θ = tan2 θ + 2 tan θ cot θ + cot2 θ • sec2 θ = tan2 θ + 2 + cot2 θ • sec2 θ – tan2 θ = 2 + cot2 θ • 1 = 2 + cot2 θ • -1 = cot2 θ

Solving a Trig Equation Linear in sin θ and cos θ • q is undefined (can’t take the square root of a negative number).

Solving Trig Equations Using a Graphing Utility • Solve 5 sin x + x = 3. Express the solution(s) rounded to two decimal places. • Put 5 sin x + x on y1 • Put 3 on y2 • Graph using the window 0 ≤ θ ≤2p • Find the intersection point(s)

Word Problems • Page 519 problem 58

Trigonometric Equations