127. If = (a, P b) is the point on the unit circle Section 2.4 θ = corresponding to , then (−a, Q is the −b) θ point on the unit circle corresponding to + π Using the graph of . θ Thus, tan(θ + π) = −b = = tan . If b = y x2 there = 3x2 1. y , vertically stretch the −a graph by a factor of 3. a exists a number π 0 < < , for which p, p tan(θ + = tan p) θ θ θ for all , then if = 0 , ( ) (0) tan = tan = p But this means that is a p multiple of π . Since no multiple of π exists in (0, the interval π) , this is impossible. Therefore, the fundamental period of π is . (θ ) θ = tan f θ π = 1 : Since tan has period , so 128. 2. = − y x2 tanθ = , reflect the graph y x2 θ does cot . Using the graph of θ cot 129. Let = (a, P b) across the be the point on the unit circle x-axis. θ corresponding to . Then cscθ = 1 = 1 θ sin b secθ = 1 = 1 cosθ a θ = = = cot 1 1 a ⎛ ⎞ b b ⎜ ⎟ θ tan ⎝ ⎠ 130. Let = (a, P b) π be the point on the unit circle θ corresponding to . Then + 2π , is any integer k k 1; 3. 2 θ = = θ tan sin b
127. If P = (a, b) is the point on the unit circle Section 2.4 corresponding to θ , then Q = (−a, −b) is the point on the unit circle corresponding to θ + π . Thus, tan(θ + π) = −b = b = tan θ . If there 1. y = 3x2 Using the graph of y = x2 , vertically stretch the exists a number −a a p, 0 < p < π , for which graph by a factor of 3. tan(θ + p) = tan θ for all θ , then if θ = 0 , tan ( p ) = tan (0) = But this means that p is a multiple of π . Since no multiple of π exists in the interval (0, π) , this is impossible. Therefore, the fundamental period of f (θ ) = tan θ is π . 128. cot θ = 1 : Since tan θ has period π , so tanθ does cot θ . 2. y = − x2 Using the graph of y = x2 , reflect the graph 129. Let P = (a, b) be the point on the unit circle across the x-axis. corresponding to θ . Then cscθ = 1 = 1 b sin θ secθ = 1 = 1 a cosθ cot θ = a = 1 = 1 b ⎛ b ⎞ ⎜ ⎟ ⎝ ⎠ tan θ 130. Let P = (a, b) be the point on the unit circle π corresponding to θ . Then 3. 1; + 2π k , k is any integer 2 tan θ = b = sin θ
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