12 ⎝ ⎠ θ θ = , cscθ = 1 = 1 = 1⋅ 2 ⋅ 3 = 2 13 43. 3 in quadrant II sin θ sin ⎛ 3 ⎞ 3 3 3 2 2 ⎜ 2 ⎟ θ θ + cos = 1 Solve for cosθ : sin ⎝ ⎠ θ = − θ = = = = cos2 1 sin secθ 1 1 1⋅ 2 2 θ =± − cosθ 1 sin 2 2 ⎜ ⎟ cosθ ⎝ ⎠ θ Since is in quadrant II, cosθ < 0 . ⎛ 1 ⎞ 1 θ = = = ⋅ = cot 1 1 1 3 θ cosθ =− 1 − sin 2 3 169 169 13 θ tan ⎛ 12 ⎞2 = − 1 − ⎜ ⎟ = 1 − 144 = − 3 3 3 3 − ⎝ 13 ⎠ 25 = − 5 ⎞ , cosθ = 2 2 13 41. 12 3 3 ⎛ 12 ⎛ 12 1 ⎞ ⎜ 13 θ θ sin sin ⎟ θ = = ⎝ ⎠ tan ⎟ = − = ⋅ ⎜ − = − 1 ⋅ 3 ⋅ 2 = − ⎛ 5 ⎞ ⎛ − 1 ⎞ θ tan = sin 13 ⎝ 2 ⎜ 3 ⎟ ⎜ − 5 ⎠ θ 13 ⎟ 5 cosθ cosθ 1 1 13 ⎛ ⎞ 2 2 3 2 2 2 4 ⎜ ⎟ 3 cscθ = = = θ ⎝ ⎠ sin ⎛ ⎞ 12 12 ⎜ ⎟ ⎝ ⎠
⎝ ⎠ 12 cscθ = 1 = 1 = 1⋅ 2 ⋅ 3 = 2 3 43. sin θ = , θ 13 in quadrant II sin θ ⎛ 3 ⎞ 3 3 3 2 2 ⎜ 2 ⎟ Solve for cosθ : sin θ + cos θ = 1 ⎝ ⎠ secθ = 1 = 1 = 1⋅ 2 = 2 cos2 θ = 1 − sin 2 θ cosθ =± 1 − sin 2 θ cosθ ⎛ 1 ⎞ 1 ⎜ ⎟ ⎝ ⎠ Since θ is in quadrant II, cosθ < 0 . cot θ = 1 = 1 = 1 ⋅ 3 = 3 cosθ =− 1 − sin 2 θ tan θ 3 3 3 3 ⎛ 12 ⎞2 = − 1 − ⎜ ⎟ = − ⎝ 13 ⎠ 1 − 144 = − 25 = − 5 169 169 13 41. sin θ 1 , cosθ = 2 2 3 3 sin θ ⎛ 12 ⎞ ⎜ 13 ⎟ 12 ⎛ 13 ⎞ 12 tan θ = = ⎝ ⎠ = ⋅ ⎜ − ⎟ = − tan θ = sin θ ⎛ − 1 ⎞ ⎜ 3 ⎟ = − 1 ⋅ 3 ⋅ 2 = − 2 cosθ ⎛ 5 ⎞ ⎜ − 13 ⎟ 13 ⎝ 5 ⎠ 5 cosθ ⎛ 2 2 ⎞ 3 2 2 2 4 1 1 13 ⎜ 3 ⎟ cscθ = = = ⎝ ⎠ sin θ ⎛ 12 ⎞ 12 ⎜ ⎟ ⎝ ⎠
62514-0132318806_ISM_02-1.docx