Stitz-Zeager_College_Algebra_e-book

Again we leave the details to the reader theorem 1015

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Unformatted text preview: coterminal with π , we find θ = π + 2πk for integers k .9 Proceeding similarly for the Quadrant 3 3 1 π IV case, we find the solution to cos(θ) = 2 here is 53 , so our answer in this Quadrant is 5π θ = 3 + 2πk for integers k . 2. If sin(θ) = − 1 , then when θ is plotted in standard position, its terminal side intersects the 2 1 Unit Circle at y = − 2 . From this, we determine θ is a Quadrant III or Quadrant IV angle π with reference angle 6 . 9 Recall in Section 10.1, two angles in radian measure are coterminal if and only if they differ by an integer multiple of 2π . Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2, . . . . 624 Foundations of Trigonometry y y 1 1 x π 6 π 6 1 −1 2 x 1 1 −2 π In Quadrant III, one solution is 76 , so we capture all Quadrant III solutions by adding integer π multiples of 2π : θ = 76 + 2πk . In Quadrant IV, one solution is 11π so all the solutions here 6 11π are of the form θ = 6 + 2πk for integers k . 3. The angles with cos(...
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