θ θ 46. tan = − 2 and 3π < < 2π (quadrant 48. = −3cos(2x) y IV) 2 The graph of vertically by is = cos y x Using the Pythagorean Identities: θ θ a factor of 3, reflected across the and sec2 = tan + 1 x-axis, 2 (−2)2 compressed horizontally by a factor of 1 . θ = + = + sec2 1 4 1 2 = =± 5 secθ 5 θ Note that secθ must be positive since lies in quadrant IV. Thus, secθ = 5 . cosθ = 1 = 1 ⋅ 5 = 5 secθ 5 5 5 θ θ = tan sin cosθ , so ⎛ ⎞ 5 2 5 sin ( ) (cosθ ) θ θ = tan = −2 ⎜ ⎟ = − . ⎝ 5 ⎠ 5 ⎛ π ⎞ cscθ = 1 = 1 =− =− 49. = − 2 cos ⎜ + ⎟ y x ⎝ ⎠ θ sin − 2 5 2 5 2 = cos The graph of y x units to the π is shifted 5 2 θ cot = 1 = 1 = − left, stretched vertically by a factor of 2, and 1 θ −2 2 reflected across the x-axis. = 2 sin(4 47. y x) = sin is stretched vertically by The graph of y x a factor of 2 and compressed horizontally by a factor of 1 . 4 50. = 3sin( − π) y x The graph of = π sin is shifted units to the y x right, and stretched vertically by a factor of 3.
46. tan θ = − 2 and 3π < θ < 2π (quadrant IV) 48. y = −3cos(2x) 2 The graph of y = cos x is vertically by Using the Pythagorean Identities: sec2 θ = tan 2 θ + 1 sec2 θ = (−2)2 + 1 = 4 + 1 = 5 secθ =± 5 Note that secθ must be positive since θ lies in a factor of 3, reflected across the x-axis, and compressed horizontally by a factor of 1 . 2 quadrant IV. Thus, secθ = 5 . cosθ = 1 = 1 ⋅ 5 = 5 secθ tan θ = sin θ cosθ 5 5 5 , so ⎛ 5 ⎞ 2 5 sin θ = ( tan θ ) (cosθ ) = −2 ⎜ ⎟ = − . ⎝ 5 ⎠ 5 ⎛ π ⎞ cscθ = 1 = 1 =− =− 49. y = − 2 cos ⎜ x + ⎟ ⎝ ⎠ sin θ − 2 5 2 5 2 5 The graph of y = cos x is shifted π 2 units to the cot θ = 1 = 1 = − 1 left, stretched vertically by a factor of 2, and 47. θ y = 2 sin(4 x) The graph of −2 2 y = sin x is stretched vertically by reflected across the x-axis. a factor of 2 and compressed horizontally by a factor of 1 . 4 50. y = 3sin( x − π) The graph of y = sin x is shifted π units to the right, and stretched vertically by a factor of 3.