− − ⎟ = cos = ⋅ ⎜ 12 13 12 12 ⎜ ⎟ 5 θ ⎞ ⎝ ⎠ ⎛ − ⎝ ⎠ ⎜ 13 ⎟ = = = secθ 1 1 5 ⎝ ⎠ 1 1 13 ⎝ ⎠ θ ⎛ sin − cosθ ⎜ 5 ⎞ 5 cscθ = − 13 ⎟ = = ⎛ ⎞ 3 3 ⎜ ⎟ θ = = = − cot 1 1 ⎝ ⎠ 1 1 13 3 − θ tan 12 ⎞ 12 secθ = = = ⎛ − 4 ⎞ 4 ⎛ ⎜ ⎟ 3 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ 13 45. θ in quadrant III θ cot = 1 = 1 = 12 cosθ 4 ⎞ 5 5 θ θ sin + cos2 = 1 ⎜ ⎟ 2 ⎝ 12 ⎠ θ θ Solve for sin : sin = 1 − cos2 5 2 θ ⎛ 5 θ tan 90º θ < θ =± − sin 1 cos2 in quadrant II sin < θ θ 180º = θ , 47. 13 , θ Since θ θ θ is in quadrant III, sin < 0 Solve for cosθ : sin2 + cos2 = 1 2 2 . θ = 1 − sin θ sin =− 1 − cos2 cos 2 θ θ θ =± 1 − sin =− 1 − ⎛ is in quadrant II, cosθ < 0 . 4 ⎞2 θ Since ⎟ − =− 1 − 16 = − 9 = − 2 ⎠ ⎝ 3 25 25 5 ⎛ 5 ⎞ θ tan = sin ⎜ ⎟ θ cosθ =− 1 − sin = ⎛ − ⎞ 2 3 ⎝ 13 ⎠ θ − = ⎝ 5 ⎠ = − 3 ⋅ − = cosθ 25 144 12 ⎜ 5 4 ⎛ 4 ⎞ ⎟ 4 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ =− 1 − = − = − 169 169 13
cos = − ⋅ ⎜ − ⎟ = 12 13 12 12 ⎜ 5 ⎟ θ ⎛ − ⎞ ⎝ ⎠ ⎝ ⎠ ⎜ 13 ⎟ secθ = 1 = 1 = 5 ⎝ ⎠ 1 1 13 cosθ ⎛ 3 ⎞ 3 ⎜ ⎟ ⎝ ⎠ cscθ = = sin θ ⎛ ⎜ = − 5 ⎞ 5 − 13 ⎟ cot θ = 1 = 1 = − 3 ⎝ ⎠ 1 1 13 tan θ ⎛ − 4 ⎞ 4 secθ = = ⎛ = − 12 ⎞ 12 ⎜ 3 ⎟ ⎜ − ⎟ ⎝ ⎠ ⎝ 13 ⎠ 45. cosθ 4 θ in quadrant III cot θ = 1 = 1 = 12 5 Solve for sin θ : sin 2 θ + cos2 θ = 1 sin 2 θ = 1 − cos2 θ tan θ 5 ⎛ 5 ⎞ 5 ⎜ ⎟ ⎝ 12 ⎠ sin θ =± 1 − cos2 θ 47. sin θ = , 90º < θ < 180º , θ 13 in quadrant II Since θ is in quadrant III, sin θ < 0 . Solve for cosθ : sin2 θ + cos2 θ = 1 2 2 sin θ =− 1 − cos2 θ cos θ = 1 − sin θ 2 =− 1 − ⎛ − ⎝ 4 ⎞2 ⎟ ⎠ =− 1 − 16 = − 9 = − 3 25 25 5 Since θ =± 1 − sin θ is in quadrant II, cosθ < 0 . 2 tan θ = sin θ ⎛ − 3 ⎞ = ⎝ 5 ⎠ = − 3 ⋅ − = cosθ =− 1 − sin 2 θ = − ⎛ 5 ⎞ ⎜ ⎟ ⎝ 13 ⎠ cosθ ⎛ 4 ⎞ 5 ⎜ 4 ⎟ 4 25 144 12 ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ =− 1 − = − = − 169 169 13