= θ < θ 2 , tan 0, so is in quadrant II 51. 4 = − 4 θ secθ = 1 = 1 sin 3 cosθ ⎛ 1 ⎞ 1 ⎜ ⎟ θ θ + = Solve for cosθ : sin2 cos2 1 ⎝ 4 ⎠ θ cos2 = 1 − sin 2 θ cot = 1 = 1 ⋅ 15 = 15 θ cosθ =± 1 − sin 2 tan 15 15 15 θ θ θ Since = θ < θ 53. secθ 2, sin 0, so is in quadrant IV < is in quadrant II, cosθ 0 . cosθ =− 1 − sin 2 θ Solve for cosθ : cosθ = 1 = 1 ⎛ ⎞2 2 secθ 2 4 5 5 =− 1 − ⎜ ⎟ ⎝ ⎠ = − − = − =− 3 1 θ θ θ 9 9 3 + = Solve for sin : sin cos2 1 2 2 2 − θ 1 cos ⎛ 2 ⎞ ⎜ ⎟ θ 2 = sin θ tan = sin θ θ θ = ⎝ 3 ⎠ = 2 ⋅ ⎛ − 3 ⎞ = − 2 5 sin =± 1 − cos cosθ θ θ < Since is in quadrant IV, sin 0 . ⎛ ⎞ ⎜ ⎟ 5 3 5 5 ⎝ ⎠ ⎜ − 3 ⎟ θ θ ⎝ ⎠ =− − sin 1 cos 1 3 3 ⎛ 1 ⎞2 = = = cscθ 1 1 3 ⎜ ⎟ θ sin ⎝ ⎠ =− − ⎜ ⎟ = − − =− =− 1 1 ⎝ 2 ⎠ 4 4 2 ⎛ 2 ⎞ 2 =− ⎜= − 3 2 ⎟ = = ⎝ secθ 1 2 ⎠ 1 θ sin 3 3 5 ⎛ ⎞ 3 ⎜ ⎟ ⋅ = − 3 cosθ 2 1 = θ tan = = ⎛ ⎞ 1 cosθ ⎛ ⎞ 5 5 5 − ⎝ ⎠ ⎜ ⎟ 3 θ = = = − = − cot 1 1 5 cscθ = 1 = 1 = − 2 = − 2 3 5 θ tan ⎛ 3 ⎞ 3 3 ⎛ − 2 5 ⎞ 2 5 2 sin θ ⎜ ⎟ 5 ⎜ − ⎟
51. sin θ = 2 , tan θ < 0, so θ is in quadrant II secθ = 1 = 1 4 = − 4 3 cosθ ⎛ 1 ⎞ 1 ⎜ ⎟ Solve for cosθ : sin2 θ + cos2 θ = 1 cos2 θ = 1 − sin 2 θ ⎝ 4 ⎠ cot θ = 1 = 1 ⋅ 15 = 15 cosθ =± 1 − sin 2 θ tan θ 15 15 15 Since θ is in quadrant II, cosθ < 0 . 53. secθ = 2, sin θ < 0, so θ is in quadrant IV cosθ =− 1 − sin 2 θ Solve for cosθ : cosθ = 1 = 1 ⎛ 2 ⎞2 4 5 5 secθ 2 =− 1 − ⎜ ⎟ ⎝ 3 ⎠ = − 1 − = − =− 9 9 3 Solve for sin θ : sin 2 θ + cos2 θ = 1 2 2 ⎛ 2 ⎞ ⎜ ⎟ sin θ = 1 − cos θ 2 tan θ = sin θ = ⎝ 3 ⎠ = 2 ⋅ ⎛ − 3 ⎞ = − 2 5 sin θ =± 1 − cos θ cosθ ⎛ 5 ⎞ 3 ⎜ 5 ⎟ 5 Since θ is in quadrant IV, sin θ < 0 . ⎝ ⎠ ⎜ − 3 ⎟ ⎝ ⎠ sin θ =− 1 − cos θ cscθ = 1 = 1 = 3 ⎛ 1 ⎞2 1 3 3 sin θ ⎛ 2 ⎞ 2 ⎜ ⎟ ⎝ ⎠ =− 1 − ⎜ ⎟ ⎝ 2 ⎠ = − 1 − =− =− 4 4 2 secθ = 1 = 1 3 3 5 =− sin θ ⎛ 3 ⎞ ⎜= − ⎟ ⎝ 2 ⎠ 3 2 cosθ ⎛ 5 ⎞ 5 5 ⎜ ⎟ tan θ = = cosθ ⎛ 1 ⎞ = − ⋅ = − 3 2 1 ⎝ 3 ⎠ ⎜ ⎟ cot θ = 1 = 1 = − 5 = − 5 cscθ = 1 = 1 = − 2 = − 2 3 tan θ ⎛ − 2 5 ⎞ 2 5 2 sin θ ⎛ 3 ⎞ 3 3 ⎜ 5 ⎟ ⎜ − ⎟
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