θ tan = sin θ = ⎝ ⎠ = − ⋅ = − =− − = 5 3 5 cosθ 1 sin 2 θ 1 − ⎛ − 2 ⎞ − 3 ⎝ ⎠ cosθ ⎝ 3 ⎠ ⎛ 4 ⎞ ⎜ ⎟ 5 4 4 =− 5 = − 5 cscθ = 1 = 1 = − 5 9 3 θ sin ⎛ − ⎞ ⎛ − ⎞ 3 3 2 ⎜ ⎟ 5 ⎜ 3 ⎟ θ sin ⎝ ⎠ ⎝ ⎠ θ = tan secθ = 1 = 1 = ⎛ ⎞ 5 = ⎜ − ⎟ 5 cosθ ⎝ ⎠ ⎞ 5 2 5 cosθ 3 ⎠ ⎝ ⎛ 4 ⎞ 4 ⎜ ⎟ ⎛ 2 3 =− ⋅ ⎜ − ⎟ ⋅ = θ = = = − cot 1 1 ⎝ ⎠ 3 5 5 5 4 θ tan 1 1 3 ⎛ − 3 ⎞ 3 ⎜ 4 ⎟ cscθ = = =− π θ θ < < π , in quadrant sin ⎛ 2 ⎞ 2 cosθ 1 ⎜ ⎟ 49. θ II ⎝ ⎠ ⎝ ⎠ 3 2 1 1 3 5 3 5 cosθ 5 5 5 θ θ + θ = ⋅ Solve for sin : sin cos2 2 = = 1 = − − ⎛ ⎞ 5 secθ = ⎜ ⎟ θ sin = 1 − cos2 2 θ ⎝ 3 ⎠ θ sin =± 1 − cos2 θ 1 1 5 5 5 θ Since θ = = = ⋅ = cot θ θ tan ⎛ 2 5 ⎞ 2 5 5 2 is in quadrant II, sin > 0 . ⎜ 5 ⎟ ⎝ ⎠
tan θ = sin θ = ⎝ 5 ⎠ = − 3 ⋅ 5 = − 3 cosθ =− 1 − sin 2 θ = − 1 − ⎛ − 2 ⎞ cosθ ⎛ 4 ⎞ ⎜ ⎟ ⎝ ⎠ 5 4 4 =− 5 = − 5 ⎝ 3 ⎠ cscθ = 1 = 1 = − 5 9 3 sin θ ⎛ − 3 ⎞ 3 ⎛ − 2 ⎞ ⎜ 5 ⎟ sin θ ⎜ 3 ⎟ ⎝ ⎠ secθ = 1 = 1 = 5 tan θ = = cosθ ⎝ ⎠ ⎛ 5 ⎞ ⎜ − ⎟ cosθ ⎛ 4 ⎞ 4 ⎜ ⎟ ⎝ ⎠ ⎝ 2 ⎛ 3 3 ⎠ ⎞ 5 2 5 =− ⋅ ⎜ − ⎟ ⋅ = cot θ = 1 = 1 = − 4 3 ⎝ 5 ⎠ 5 5 tan θ ⎛ − 3 ⎞ 3 1 1 3 ⎜ 4 ⎟ cscθ = = =− 49. cosθ 1 ⎝ ⎠ π < θ < π , θ in quadrant II sin θ ⎛ 2 ⎞ 2 ⎜ ⎟ ⎝ ⎠ 3 2 1 1 3 5 3 5 Solve for sin θ : sin 2 θ + cos2 θ = 1 secθ = cosθ = = − ⎛ 5 ⎞ ⎜ ⎟ ⋅ = − 5 5 5 sin 2 θ = 1 − cos2 θ ⎝ 3 ⎠ sin θ =± 1 − cos2 θ 1 1 5 5 5 Since θ is in quadrant II, sin θ > 0 . cot θ = = = ⋅ = tan θ ⎛ 2 5 ⎞ 2 5 5 2 ⎜ 5 ⎟ ⎝ ⎠
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