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9-4: Using Trigonometric Identities

9-4: Using Trigonometric Identities. 9-4: Using Trig Identities. Ex 1: Use Double-Angle Identities Solve 5 cos x + 3 cos 2x = 3 You’ve got cos x and cos 2x, use double-angle identity to rewrite cos 2x in terms of cos 2 x cos 2x = 2 cos 2 x – 1 5 cos x + 3 (2 cos 2 x – 1) = 3

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9-4: Using Trigonometric Identities

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  1. 9-4: Using Trigonometric Identities

  2. 9-4: Using Trig Identities • Ex 1: Use Double-Angle Identities • Solve 5 cos x + 3 cos 2x = 3 • You’ve got cos x and cos 2x, use double-angle identity to rewrite cos 2x in terms of cos2 x • cos 2x = 2 cos2 x – 1 • 5 cos x + 3 (2 cos2 x – 1) = 3 • 5 cos x + 6 cos2 x – 3 = 3 This looks quadratic… • 6 cos2 x + 5 cos x – 6 = 0 • (3 cos x – 2)(2 cos x + 3) = 0 Factor (next slide)

  3. 9-4: Using Trig Identities • (3 cos x – 2)(2 cos x + 3) = 0 • 3 cos x – 2 = 0 2 cos x + 3 = 0 • cos x = 2/3 cos x = - 3/2 • x = cos-1 (2/3) This is a dead end (no solution)… • x ≈ 0.8411 + 2πk • 2nd solution: x ≈ 5.4421 + 2πk

  4. 9-4: Using Trig Identities • Ex 2: Use Double-Angle Identities • Solve sin x cos x = 1 • You’ve got the double angle identity 2 sin x cos x = sin 2x, which is really close to the original problem, so let’s manipulate that first. • 2 sin x cos x = sin 2x • sin x cos x = ½ sin 2x • sin x cos x = 1 • ½ sin 2x = 1 • sin 2x = 2 • No solution

  5. 9-4: Using Trig Identities • Ex 3: Use Double-Angle Identities • Find exact solutions of cos2 x – sin2 x = ½ • Pythagorean is an option, but let’s continue to use our new identities. • Because cos 2x = cos2 x – sin2 x, we can rewrite the equation • cos2 x – sin2 x = ½ • cos 2x = ½ • 2x = cos-1 (½) • 2x = π/3 + 2πk or (2nd solution) 2x = 5π/3 + 2πk • x = π/6 + πk or x = 5π/6 + πk

  6. 9-4: Using Trig Identities • Ex 4: Use Addition Identities • Find exact solutions of sin 2x cos x + cos 2x sin x = 1 • This is an addition identity: sin(x + y) = sin x cos y + cos x sin y • Let 2x be “x” and x by “y” • sin 2x cos x + cos 2x sin x = 1 • sin(2x + x) = 1 • sin 3x = 1 • 3x = sin-1 (1) • 3x = π/2 + 2πk no 2nd solution (π – π/2 = π/2) • x = π/6 + 2πk/3

  7. 9-4: Using Trig Identities • Ex 5: Use Half-Angle Identities • Find the solutions of sin x = sin x/2, where 0 < x < 2π • Note: Because we’ve got a domain, no “+ 2πk” shenanigans • sin x = sin x/2 • half-angle identity • square both sides • Pythagorean Identity

  8. 9-4: Using Trig Identities • Ex 5: Use Half-Angle Identities • continuing… • 2 – 2 cos2 x = 1 – cos x multiply by 2 • -2 cos2 x + cos x + 1 = 0 set equal to 0 • 2 cos2 x – cos x – 1 = 0 make squared term positive • (2 cos x + 1)(cos x – 1) = 0 factor (next slide)

  9. 9-4: Using Trig Identities • (2 cos x + 1)(cos x – 1) = 0 • 2 cos x + 1 = 0 cos x – 1 = 0 • cos x = -½ cos x = 1 • x = cos-1 (-½) x = cos-1 (1) • x = 2π/3 x = 0 • 2nd solution: x = 4π/3 2nd solution: not necessary (0 = 2π)

  10. 9-4: Using Trig Identities • Assignment • Page 608 – 609 • Problems 1 – 35 (odds) • Show work • Note that all answers are interval solutions (no +2πk stuff)

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