Trigonometric Equations In quadratic form using identities or

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

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

Solving a Trig Equation in Quadratic Form o o o 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 5 p/6 θ = p/2

Solving a Trig Equation in Quadratic Form o o o 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 5 p/4 θ=0

Solving a Trig Equation Using Identities o o 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 sin 2 θ Use the pythagorean identities to change sin 2 θ to cos θ

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

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

Solving a Trig Equation Using Identities o o o o cos 2 θ – sin 2 θ + sin θ = 0 1 – sin 2 θ + sin θ = 0 -2 sin 2 θ + sin θ + 1 = 0 2 sin 2 θ – sin θ – 1 = 0 Let c = sin θ 2 c 2 – c – 1 = 0 (2 c + 1) (c – 1) = 0

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

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

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

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

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

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

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

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

Solving Trig Equations Using a Graphing Utility o o o Solve 5 sin x + x = 3. Express the solution(s) rounded to two decimal places. Put 5 sin x + x on y 1 Put 3 on y 2 Graph using the window 0 ≤ θ ≤ 2 p Find the intersection point(s)

Word Problems o Page 519 problem 58