θ (−2)2 tan = 32 −1 = 9 −1 θ csc2 = 1 + = 1 + 4 2 = θ =± = ±2 8 tan 8 = 5 cscθ =± 5 2 θ Note that cscθ must be positive because lies θ θ Note that tan must be negative since lies in in quadrant II. Thus, cscθ = 5 . θ = −2 quadrant IV. Thus,. tan 2 θ sin = 1 = 1 ⋅ 5 = 5 . cscθ 5 5 5 θ cot = cosθ , so θ sin = −2 = − . cosθ = ⎞ θ sin 2 θ cot ⎛ 5 5 = = cosθ 1 1 ( ) ( ) ⎜ ⎟ secθ ⎝ ⎠ θ 5 5 tan = 1 = 1 = − , so θ θ tan = sin 1 3 cosθ −2 θ cot 2 ( ) (cosθ ) θ θ sin = tan = −2 2 ⎛ 1 ⎞ = − = = =− =− 2 2 . secθ 1 1 ⎜ ⎟ cosθ 2 5 2 2 5 cscθ = 1 = 1 = − 3 ⋅ 2 = − 3 2 θ sin 3 2 4 2 2 2 2 θ = = ⋅ = − cot 1 1 2 2 θ tan −2 2 2 4
tan 2 θ = 32 −1 = 9 −1 = 8 tan θ =± 8 = ±2 2 Note that tan θ must be negative since θ lies in quadrant IV. Thus,. tan θ = −2 2 . csc2 θ = 1 + (−2)2 = 1 + 4 = 5 cscθ =± 5 Note that cscθ must be positive because θ lies in quadrant II. Thus, cscθ = 5 . sin θ = 1 = 1 ⋅ 5 = 5 cscθ 5 5 5 cot θ = cosθ , so sin θ cosθ = cot θ sin θ ⎛ 5 ⎞ 2 5 = −2 = − . cosθ = 1 = 1 ( ) ( ) ⎜ ⎟ secθ tan θ = sin θ 3 , so tan θ = 1 = 1 = − 1 ⎝ 5 ⎠ 5 cosθ cot θ −2 2 sin θ = ( tan θ ) (cosθ ) = −2 2 ⎛ 1 ⎞ = − 2 2 . secθ = 1 = 1 =− =− ⎜ ⎟ cosθ 2 5 2 5 2 cscθ = 1 = 1 = − 3 ⋅ 2 = − 3 2 sin θ 2 2 3 2 2 2 4 cot θ = 1 = 1 ⋅ 2 = − 2 tan θ −2 2 2 4
62514-0132318806_ISM_02-1.docx