cscθ = 1 = 1 1⋅ 5 = = quadrant III. Thus, secθ 13 5 θ sin 4 4 5 4 = = = − cosθ 1 1 5 = secθ 1 = 1 1⋅ 5 = 5 = cosθ 3 3 3 secθ 13 13 1 1 3 3 , so θ tan = sin θ θ 4 4 cot = = θ ( ) (cosθ ) tan θ θ sin = tan = 12 ⎛ − 5 ⎞ = − 12 4 = = 1⋅ 4 32. ⎝ ⎠ 5 13 13 θ tan = = = = − cscθ 1 1 13 θ π θ and 0 < < , so lies in quadrant I. 1 2 Using the Pythagorean Identities: θ = θ + sec2 tan 1 2 θ sin 12 12 θ cot = 1 = 1 = 5 16 16 θ sec2 = ⎛ 1 ⎞ ⎝ ⎠ θ tan + = + = 1 1 1 17 12 12 secθ =± 17 =± 17 16 4 θ Note that secθ must be positive since lies in = quadrant I. Thus, secθ 17 . 4
cscθ = 1 = 1 = 1⋅ 5 = 5 quadrant III. Thus, secθ 13 sin θ 4 4 4 5 cosθ = 1 = 1 = − 5 secθ = 1 = 1 = 1⋅ 5 = 5 cosθ 3 3 3 secθ 13 13 1 1 3 3 tan θ = sin θ , so cot θ = = tan θ 4 = 1⋅ = 4 4 sin θ = ( tan θ ) (cosθ ) = 12 ⎛ − 5 ⎞ = − 12 32. tan θ = 1 4 and 0 < θ < π , so θ lies in quadrant I. 2 5 ⎝ cscθ = 1 = 1 = − 13 13 ⎠ 13 Using the Pythagorean Identities: sec2 θ = tan 2 θ + 1 sin θ 12 12 cot θ = 1 = 1 = 5 sec2 θ = ⎛ 1 ⎞ ⎝ ⎠ + 1 = 1 + 1 = 17 16 16 tan θ 12 12 secθ =± 17 =± 17 16 4 Note that secθ must be positive since θ lies in quadrant I. Thus, secθ = 17 . 4
62514-0132318806_ISM_02-1.docx