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標題:

trigonometric equation

發問:

1. given that tan theta = -3 and cos theta>0 find the value of sin theta+cos theta2. given that sin theta = -(開方3)/2 and tan theta<0 find cos theta and tan thetahence find (1/cos theta - tan theta)(1/cos theta+tan theta)3. what is the reference angle for the angle of rotation theta if theta is... 顯示更多 1. given that tan theta = -3 and cos theta>0 find the value of sin theta+cos theta 2. given that sin theta = -(開方3)/2 and tan theta<0 find cos theta and tan theta hence find (1/cos theta - tan theta)(1/cos theta+tan theta) 3. what is the reference angle for the angle of rotation theta if theta is -140degrees 4.find the values of sin 405degree and-tan (-3pie/4) 5. show cos[(2pie-theta)*tan(pie-theta)]/sin(pie-theta) = -1

最佳解答:

(1) tan θ = -3, then from the given: sin θ = -3/√10 and cos θ = 1/√10 Hence sin θ + cos θ = -2/√10 (2) sin θ = -√3 / 2 since tan θ < 0, cos θ > 0 and hence: cos θ = √(1 - sin2 θ) = 1/2 tan θ = -√3 Hence (1/cos θ - tan θ)(1/cos θ + tan θ) = 1/cos2 θ - tan2 θ = 1/cos2 θ - sin2 θ/cos2 θ = (1 - sin2 θ)/cos2 θ = 1 (3) θ = - 140 deg i.e. in quadrant III, ref. angle = 40 deg. (4) sin 405 = sin (360 + 45) = sin 45 = 1/√2 - tan (-3π/4) = - tan (-π + π/4) = - tan π/4 = - 1 (5) [cos (2π - θ) tan (π - θ)]/sin (π - θ) = [cos θ (- tan θ)]/sin θ = - sin θ/sin θ = - 1 2009-11-11 14:17:44 補充： Q1) If tan x = a/b, then we have: sin x = a/√(a^2 + b^2) or -a/√(a^2 + b^2) cos x = b/√(a^2 + b^2) or -b/√(a^2 + b^2)

 

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