secθ = 1 = 1 =− ⋅ 3 θ θ sin =− 1 − cos2 cosθ 1 − 16 9 = − 3 2 2 2 = − 4 ⎞2 = − − ⎛ − 1 ⎟ = − 4 ⎛ ⎞ 2 2 ⎜ ⎟ ⎝ 3 ⎠ 25 25 5 ⎝ 5 ⎠ θ cot = 1 = 1 = − 4 ⋅ 2 = − 2 2 cscθ = 1 = 1 = − 5 θ tan 3 ⎞ 3 θ sin ⎛ − ⎛ 2 ⎞ 2 2 ⎜ − ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ 4 1 , is in quadrant II θ θ = 3 , sin < 0, so is in quadrant III 55. 3 sin 4 tan θ > θ 0, 57. θ tan θ so θ = + θ Solve for secθ : sec2 1 tan 2 θ θ Solve for secθ : sec2 = 1 + tan 2 secθ =± 1 + tan 2 =± + θ secθ 1 tan θ 2 θ Since θ < Since is in quadrant II, secθ 0 . is in quadrant III, secθ < 0 . 2 secθ =− 1 + tan 2 θ secθ =− 1 + tan θ 2 ⎛ ⎞2 3 =− 1 + ⎛ − 1 ⎞ = − 1 + 1 = 10 = − 10 9 25 5 − 9 3 =− + ⎜ ⎟ 1 = − 1 + = − ⎜ 3 ⎟ =− 9 ⎝ 4 ⎠ ⎝ ⎠ 16 16 4 cosθ = 1 = − 4 3 3 10 = = =− =− cosθ ⎜ − secθ 5 ⎟ ⎛ 10 10 10 ⎞ secθ θ =− − sin 1 cos2 θ ⎝ 3 ⎠ =− − = − = − 1 16 9 =− − ⎛ 1 ⎞2 4 ⎟ θ θ sin 1 − cos2 3 − = = 1 − ⎜ − ⎝ 5 ⎠ 25 25 5 2 ⎛ 3 10 ⎞ ⎟ = 1 − 90 = = = − cscθ 1 1 ⎝ ⎠ 10 100 5 θ sin ⎛ − 3 ⎞ 3
secθ = 1 = 1 =− ⋅ 3 sin θ =− 1 − cos2 θ cosθ ⎛ 2 2 ⎞ ⎜ ⎟ 2 2 2 4 = − 1 − ⎛ − 4 ⎞2 ⎟ = − 1 − 16 = − 9 = − 3 ⎝ 3 ⎠ ⎝ 5 ⎠ 25 25 5 cot θ = 1 = 1 = − 4 ⋅ 2 = − 2 2 cscθ = 1 = 1 = − 5 tan θ ⎛ 2 ⎞ 2 2 sin θ ⎛ − 3 ⎞ 3 ⎜ − ⎟ ⎜ 5 ⎟ ⎝ 4 ⎠ ⎝ ⎠ 1 55. tan θ = 3 , sin θ < 0, so θ is in quadrant III 4 57. tan θ , sin θ > 0, so θ is in quadrant II 3 Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Solve for secθ : sec2 θ = 1 + tan 2 θ secθ =± 1 + tan 2 θ Since θ is in quadrant III, secθ < 0 . Since θ is in quadrant II, secθ < 0 . 2 secθ =− 1 + tan 2 θ secθ =− 1 + tan θ 2 ⎛ 3 ⎞2 9 25 5 =− 1 + ⎛ − 1 ⎞ = − 1 + 1 = − 10 = − 10 =− 1 + ⎜ ⎟ = − 1 + = − =− ⎜ 3 ⎟ 9 9 3 ⎝ 4 ⎠ cosθ = 1 = − 4 16 16 4 ⎝ ⎠ 3 3 10 cosθ = = =− =− secθ 5 secθ ⎛ 10 ⎞ ⎜ − ⎟ 10 10 sin θ =− 1 − cos2 θ ⎝ 3 ⎠ =− 1 − ⎛ − 4 ⎞2 ⎟ =− 1 − 16 = − 9 = − 3 sin θ = 1 − cos2 θ ⎝ 5 ⎠ 25 25 5 ⎛ = 1 − ⎜ − 2 3 10 ⎞ ⎟ = 1 − 90 cscθ = 1 = 1 = − 5 ⎝ 10 ⎠ 100 sin θ ⎛ − 3 ⎞ 3
62514-0132318806_ISM_02-1.docx