⎝ ⎠ ⎝ 2 ⎠ cosθ 52. θ θ > cot = 1 = 1 = − 3 1 , tan 0 =− 4 θ tan − 3 3 θ = Since tan sin θ > 0 and cosθ < 0 , sin < 0 . θ cosθ θ θ θ Solve for sin : sin + cos2 = = θ < θ 54. cscθ 3, cot 0, so is in quadrant 2 1 II θ sin =± 1 − cos2 θ θ Solve for sin : sin = 1 = 1 cscθ θ 3 θ θ sin =− 1 − cos2 θ θ Solve for cosθ : sin2 + cos2 = 1 ⎟ = − − θ 1 sin ⎠ 2 Since =− − ⎛ 1 is in quadrant II, cosθ < 0 . θ 2 − 1 − 1 ⎝ 16 ⎞2 1 =− 15 = − 15 =− − θ cosθ 1 sin 16 4 ⎛ 15 ⎞ ⎛ 1 ⎞2 ⎜ − ⎟ =− 1 − ⎜ ⎟ = − − = − 1 1 8 2 2 ⎝ ⎠ 3 9 9 3 θ tan = sin ⎛ 4 ⎞ ⎛ 1 ⎞ θ = =− ⋅ − = 15 15 cosθ ⎛ ⎞ 1 ⎜ ⎟ ⎜ ⎟ 4 1 ⎜ − ⎟ ⎝ ⎠ = ⎝ ⎠ 3 θ = θ tan sin ⎝ 4 ⎠ cosθ ⎛ ⎞ 2 2 ⎜ ⎟ cscθ = 1 = 1 = − 4 ⋅ 15 = − 4 ⎝ ⎠ 3 15 ⎜ ⎟ θ sin 15 15 15 = 1 ⋅ ⎛ − 3 ⎞ ⋅ 2 = − 2 ⎛ ⎞ 15 ⎝ 4 ⎠ 3 ⎝ 2 2 ⎠ 2 4
⎝ ⎠ ⎝ 2 ⎠ 52. cosθ 1 , tan θ > 0 cot θ = 1 = 1 = − 3 =− 4 Since tan θ = sin θ > 0 and cosθ < 0 , sin θ < 0 . tan θ − 3 3 cosθ Solve for sin θ : sin 2 θ + cos2 θ = 1 sin θ =± 1 − cos2 θ sin θ =− 1 − cos2 θ 54. cscθ = 3, cot θ < 0, so θ is in quadrant II Solve for sin θ : sin θ = 1 = 1 cscθ 3 Solve for cosθ : sin2 θ + cos2 θ = 1 =− 1 − ⎛ − ⎝ 1 ⎞2 ⎟ = − ⎠ 1 − 1 16 Since θ 1 − sin 2 θ is in quadrant II, cosθ < 0 . 2 =− 15 = − 15 cosθ =− 1 − sin θ 16 4 ⎛ 15 ⎞ ⎜ − ⎟ ⎛ 1 ⎞2 =− 1 − ⎜ ⎟ = − ⎝ 3 ⎠ 1 − 1 = − 8 2 2 9 9 3 tan θ = sin θ 15 ⎛ 4 ⎞ = =− ⋅ − = 15 ⎛ 1 ⎞ cosθ ⎛ 1 ⎞ 4 ⎜ 1 ⎟ ⎜ ⎟ ⎜ − ⎟ ⎝ ⎠ tan θ = sin θ = ⎝ 3 ⎠ ⎝ 4 ⎠ cosθ ⎛ 2 2 ⎞ ⎜ ⎟ cscθ = 1 = 1 = − 4 ⋅ 15 = − 4 15 ⎝ 3 ⎠ sin θ ⎛ 15 ⎞ ⎜ ⎟ 15 15 15 = 1 ⋅ ⎛ − 3 ⎞ ⋅ 2 = − 2 ⎝ 4 ⎠ 3 ⎝ 2 2 ⎠ 2 4
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