(0 ) (0) θ If = 0 , then cos + = cos . But p 121. Let = ( be the point on the unit circle that P x, y) (π) (0) = π . Thus cos = −1 = cos = 1 , or p −1 = 1 . This is impossible. The smallest ⎛ π + ⎞ = ⎛ π ⎞ = = π then cos cos 0 ; so that . p p ⎜ ⎟ ⎜ ⎟ positive number for which cos(θ + = cosθ p p) ⎝ ⎠ ⎝ ⎠ corresponds to an angle Consider the equation t. θ for all must then be = 2π . p tan = = . Then = . Now + = 1 t y a y ax x2 y2 , x Since cosθ has period 2π , so and cosθ so + = 1 . Thus, =± x2 a2 x2 x does secθ . 1 + a2 = 1 : a =± . That is, for any real number y a , 1 + a2 there is a point = ( on the unit circle for P x, y) which tan = . In other words, t a −∞ < tan < ∞ , and the range of the t tangent function is the set of all real numbers. 126. θ θ cscθ = 1 : Since sin has period 2π , so sin does cscθ .
121. Let P = ( x, y) be the point on the unit circle that then cos ⎛ π + p ⎞ = cos ⎛ π ⎞ = 0 ; so that p = π . ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ If θ = 0 , then cos (0 + p ) = cos (0) . But p = π . Thus cos (π) = −1 = cos (0) = 1 , or −1 = 1 . This is impossible. The smallest positive number p for which cos(θ + p) = cosθ corresponds to an angle t. Consider the equation tan t = y = a . Then y = ax . Now x2 + y2 = 1 , for all θ must then be p = 2π . x so x2 + a2 x2 = 1 . Thus, x =± 1 + a2 and = 1 : Since cosθ has period 2π , so cosθ does secθ . a y =± . That is, for any real number a , 1 + a2 there is a point P = ( x, y) on the unit circle for which tan t = a . In other words, −∞ < tan t < ∞ , and the range of the tangent function is the set of all real numbers. 126. cscθ = 1 : Since sin θ has period 2π , so sin θ does cscθ .
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