⎛ 5 ⎞ θ tan = sin θ − θ sin 1 cos2 ⎜ ⎟ = ⎝ ⎠ 13 θ = 5 ⋅ ⎛ − 13 ⎞ = − = 2 5 cosθ ⎜ ⎟ = − ⎛ − − = = 13 12 1 1 1 8 2 ⎛ 12 ⎞ 12 1 ⎞ = 2 ⎜ − ⎟ ⎝ ⎠ 9 9 3 ⎜ 3 ⎟ ⎝ 13 ⎠ ⎝ ⎠ cscθ = 1 = 1 = 13 ⎛ 2 2 ⎞ ⎜ ⎟ θ sin θ sin ⎝ 3 2 2 ⎛ ⎛ 5 ⎞ 5 ⎠ 3 ⎞ ⎜ 13 ⎟ θ = = = ⎟ = − tan 2 2 ⋅ ⎜ − ⎝ ⎠ cosθ ⎝ 3 ⎛ 1 ⎞ ⎜ ⎟ ⎠ 1 = = = − secθ 1 1 ⎝ ⎠ 3 13 cosθ ⎛ − 12 ⎞ 12 cscθ = 1 = 1 = 3 ⋅ 2 = 3 2 ⎜ 13 ⎟ θ ⎝ ⎠ sin 2 2 2 4 ⎛ 2 2 ⎞ ⎜ ⎟ θ = = = − cot 1 1 12 ⎝ 3 ⎠ θ tan 1 1 ⎛ − ⎞ 5 5 ⎜ ⎟ 12 1 secθ = = = −3 ⎝ ⎠ cosθ ⎛ ⎞ ⎜ − ⎟ ⎝ ⎠ θ θ = 4 , 270º < < 360º ; cot = 1 = 1 ⋅ 2 = − 2 48. 5 cosθ θ in quadrant IV θ θ θ Solve for sin : sin + cos2 = 2 tan − 2 2 2 4 1 θ 2 3π θ sin =± 1 − cos2 θ θ sin π < < III θ θ =− 50. , in , θ Since 3 2 θ is in quadrant IV, sin < 0 θ θ Solve for cosθ : sin + cos2 = 2 . 1 2 = − = θ θ θ θ Since =± 1 − sin sin =− 1 − cos2 ⎛ ⎞ = − 1 − 4 − 2 ⎝ ⎠ 25 5 is in quadrant III, cosθ < 0 . ⎛ − 3 ⎞ 2 9 3
tan θ = sin θ ⎛ 5 ⎞ ⎜ ⎟ = ⎝ 13 ⎠ = 5 ⋅ ⎛ − 13 ⎞ = − 5 sin θ = 1 − cos2 θ 2 cosθ ⎛ 12 ⎞ 13 ⎜ 12 ⎟ 12 = 1 − ⎛ − 1 ⎞ = 1 − 1 = 8 = 2 2 ⎜ − ⎟ ⎝ ⎠ ⎜ 3 ⎟ 9 9 3 ⎝ 13 ⎠ ⎝ ⎠ cscθ = 1 = 1 = 13 ⎛ 2 2 ⎞ ⎜ ⎟ sin θ ⎛ 5 ⎞ 5 sin θ ⎝ 3 ⎠ 2 2 ⎛ 3 ⎞ ⎜ 13 ⎟ tan θ = = = ⋅ ⎜ − ⎟ = − 2 2 ⎝ ⎠ cosθ ⎛ 1 ⎞ ⎜ ⎟ 3 ⎝ 1 ⎠ secθ = 1 = 1 = − 13 ⎝ 3 ⎠ cosθ ⎛ − 12 ⎞ 12 cscθ = 1 = 1 = 3 ⋅ 2 = 3 2 ⎜ 13 ⎟ ⎝ ⎠ sin θ ⎛ 2 2 ⎞ ⎜ ⎟ 2 2 2 4 cot θ = 1 = 1 = − 12 ⎝ 3 ⎠ tan θ ⎛ − 5 ⎞ 5 1 1 ⎜ 12 ⎟ secθ = = = −3 1 ⎝ ⎠ cosθ ⎛ ⎞ ⎜ − ⎟ ⎝ ⎠ 48. cosθ = 4 , 270º < θ < 360º ; 5 θ in quadrant IV cot θ = 1 = 1 ⋅ 2 = − 2 Solve for sin θ : sin 2 θ + cos2 θ = 1 tan θ − 2 2 2 4 2 3π sin θ =± 1 − cos2 θ 50. sin θ =− , π < θ < , θ in III Since θ is in quadrant IV, sin θ < 0 . 2 3 2 Solve for cosθ : sin 2 θ + cos2 θ = 1 sin θ =− 1 − cos2 θ = − 1 − ⎛ 4 ⎞ ⎝ ⎠ ⎛ − 3 ⎞ 9 3 = − = − 25 5 Since θ =± 1 − sin 2 θ is in quadrant III, cosθ < 0 . 2
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