secθ = 1 = 1 = − 13 = = = − cscθ 1 1 5 cosθ θ ⎛ − ⎞ ⎛ − ⎞ 5 5 sin 3 3 ⎜ 13 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ θ cot = 1 = 1 = − secθ = 1 = 1 = − 5 5 θ tan ⎛ − 12 ⎞ 12 cosθ ⎛ − 4 ⎞ 4 ⎜ 5 ⎟ ⎜ ⎟ 5 ⎝ ⎠ ⎝ ⎠ θ cot = 1 = 1 = 4 5 44. ⎛ 3 ⎞ 3 tan cosθ = 3 ⎜ ⎟ θ ⎝ ⎠ in quadrant IV θ , θ θ + θ = Solve for sin : sin cos2 1 2 θ = − sin 1 cos2 2 sin θ θ in 5 46. =− θ , III θ θ =± 1 − cos2 θ θ = − cos2 1 sin 2 θ < θ θ is in quadrant IV, sin 0 . sin + cos2 θ 2 Since 2 = 1 2 Solve for cosθ : ⎝ 5 ⎠ θ ± 1 − sin θ θ sin =− 1 − cos2 = is in quadrant III, cosθ < 0 . 16 − 4 θ Since 2 =− ⎛ 3 ⎞ 25 ⎜ ⎟ 5 θ =− − = − cosθ 1 sin 2 1 − ⎛ − 5 ⎞ ⎛ − 4 ⎞ ⎜ 13 ⎟ θ tan = sin ⎝ ⎠ 144 12 = ⎝ ⎠ = − ⋅ = − θ 5 4 5 4 ⎝ ⎠ 169 13 cosθ ⎛ − 5 ⎞ =− 5 3 ⎛ 3 ⎞ =− ⎜ ⎟ 3 5 ⎛ cscθ = 1 = θ sin 13 ⎞ 1 = − ⎜ 13 ⎟ 5 5 θ sin θ tan = = ⎛ − 4 ⎞ 4
secθ = 1 = 1 = − 13 cscθ = 1 = 1 = − 5 cosθ ⎛ − 5 ⎞ 5 sin θ ⎛ − 3 ⎞ 3 ⎜ 13 ⎟ ⎜ 5 ⎟ ⎝ ⎠ cot θ = 1 = 1 = − 5 ⎝ ⎠ secθ = 1 = 1 = − 5 tan θ ⎛ − 12 ⎞ 12 cosθ ⎛ − 4 ⎞ 4 ⎜ 5 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ cot θ = 1 = 1 = 4 44. cosθ = 3 , 5 θ in quadrant IV tan θ ⎛ 3 ⎞ 3 ⎜ ⎟ ⎝ ⎠ Solve for sin θ : sin 2 θ + cos2 θ = 1 sin 2 θ = 1 − cos2 θ 46. sin θ 5 =− , θ in III Since θ θ =± 1 − cos2 θ is in quadrant IV, sin θ < 0 . 2 Solve for cosθ : sin 2 θ + cos2 θ = 1 cos2 θ = 1 − sin 2 θ 2 sin θ =− 1 − cos2 θ = − ⎛ 3 ⎞ ⎜ ⎟ ⎝ 5 ⎠ 16 4 =− 25 5 Since θ 1 − sin θ is in quadrant III, cosθ < 0 . 2 cosθ =− 1 − sin 2 θ = − 1 − ⎛ − 5 ⎞ ⎛ − 4 ⎞ ⎜ 13 ⎟ tan θ = sin θ = ⎝ 5 ⎠ = − 4 ⋅ 5 = − 4 ⎝ ⎠ 144 12 cosθ ⎛ 3 ⎞ ⎜ ⎟ ⎝ ⎠ 5 3 3 =− =− 169 13 ⎛ − 5 ⎞ cscθ = 1 = 1 = − 5 sin θ ⎜ 13 ⎟ 5 ⎛ 13 ⎞ 5 sin θ ⎛ − 4 ⎞ 4 tan θ = = ±
62514-0132318806_ISM_02-1.docx