University:
Winston-Salem Forsyth County Schools
Course:
AP Calculus AB
Academic year:

2014
Views:

321

Pages:

4
Author:

Vasientep9m

5.5 Notes Double Angle Formulas sin 2u = 2 sin u cos u cos 2u = cos^2 u - sin^2 u = 2 cos^2 u - 1 = 1 - 2 sin^2 u Find the following exact values: sin 2θ, cos 2θ, tan 2θ sin 2θ = 2sinθ cosθ = 2(8/17)(15/17) cos 2θ = cos^2θ - sin^2θ = (15/17)^2 - (8/17)^2 sin 2θ = 240/289 cos 2θ = 161/289 tan 2θ = 2tanθ / (1 - tan^2θ) = 2(8/15) / (1 - (8/15)^2) = 240/161 OR 240/289 / 161/289 = 240/161 Given: cos u = -3/5 = x/r π/2 < u < π Quad II x = -3 r = 5 y = 4 Find the following exact values: sin 2u, cos 2u, tan 2u sin 2u = 2sinu cosu = 2(4/5)(-3/5) = -24/25 cos 2u = cos^2u - sin^2u = (3/5)^2 - (4/5)^2 = -7/25 tan 2u = 24/7 what quadrant is 2u in? Quad III ble sinau & cosdu are negative II. Solve on [0, 2π) Ex: 2sinθcosθ = -1 work backwards sin2θ = 1 2θ = π/2 + 2πn θ = π/4 + πn θ = π/4, 7π/4 Ex: cos²θ - sin²θ = 1/2 [cos2θ = 1/2] [2θ = π/3 + 2πn] [2θ = 5π/3 + 2πn] [θ = π/6 + πn] [θ = 5π/6 + πn] 4 solutions. θ = π/6, 7π/6, 5π/6, 11π/6 sinθ - (1-2sin²θ) = 0 sinθ - 1 + 2sin²θ = 0 2sin²θ + sinθ - 1 = 0 factor (2sinθ - 1)(sinθ + 1) = 0 sinθ = 1/2 θ = π/6, 5π/6 sinθ = -1 θ = 3π/2 Ex: sin2θ = sinθ 2sinθcosθ = sinθ 2sinθcosθ - sinθ = 0 sinθ(2cosθ - 1) = 0 sinθ = 0 θ = 0, π cosθ = 1/2 θ = π/3, 5π/3 WORK backwards Ex: sinθ - cos2θ = 0 use parentheses!! Half-Angle Formulas sin(u/2) = ±√((1-cos u)/2) cos(u/2) = ±√((1+cos u)/2) tan(u/2) = (1-cos u)/sin u or sin u/(1+cos u) the sign depends on which quadrant u/2 is in III. Find the sin(θ/2), cos(θ/2), tan(θ/2) cos θ = 3/5 sin(θ/2) = √((1-3/5)/2) = √((2/5)/2) = √(1/5) = 1/√5 = √5/5 cos(θ/2) = √((1+3/5)/2) = √((8/5)/2) = √(4/5) = 2/√5 = 2√5/5 tan(θ/2) = (1-cos θ)/sin θ = (1-3/5)/(4/5) = (2/5)/(4/5) = 1/2 IV. Find the sine, cosine, and tangent of 157°30' u = 315° Quad II sin(u/2) = +√((1-cos u)/2) = +√((1-(-√2/2))/2) = +√((2+√2)/4) = √(2+√2)/2 cos(u/2) = +√((1+cos u)/2) = +√((1+(-√2/2))/2) = √((2-√2)/4) = √(2-√2)/2 tan(u/2) = (1-cos u)/sin u = (1-(-√2/2))/(√2/2) = (2+√2)/√2 = (2+√2)√2/2 = 2+√2 V. Find sin(θ/2), cos(θ/2), tan(θ/2) given the following: tan θ = -12/5 3π/2 < θ < 2π Quad IV sin θ = -12/13 cos θ = 5/13 r = 13 sin(θ/2) = +√((1-cos θ)/2) = +√((1-5/13)/2) = +√((8/13)/2) = +√(4/13) = 2/√13 = 2√13/13 cos(θ/2) = +√((1+cos θ)/2) = +√((1+5/13)/2) = +√((18/13)/2) = +√(9/13) = 3/√13 = 3√13/13 tan(θ/2) = (1-cos θ)/sin θ = (1-5/13)/(-12/13) = (8/13)/(-12/13) = -8/12 = -2/3 check VI. Solve for the zeros. f(x) = sin(x/2) + cos x - 1 0 = sin(π/6) + cos(π/3) - 1 0 = 1/2 + 1/2 - 1 0 = sin(x/2) + cos x - 1 0 = ±√((1-cos x)/2) + cos x - 1 1 - cos x = ±√((1-cos x)/2) (1 - cos x)^2 = (1-cos x)/2 2 - 4cos x + 2cos^2 x = 1 - cos x 2cos^2 x - 3cos x + 1 = 0 cos x = 1/2 x = π/3, 5π/3 Check Solutions. Anytime you square both sides