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Since is a quadrant ii angle 13 cos 12 we now set

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Unformatted text preview: θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y -axis. y y 1 1 π 2 x x 1 1 π 2 While, technically speaking, π isn’t a reference angle we can nonetheless use it to find our 2 answers. If we follow the procedure set forth in the previous examples, we find θ = π + 2πk 2 π and θ = 32 + 2πk for integers, k . While this solution is correct, it can be shortened to θ = π + πk for integers k . (Can you see why this works from the diagram?) 2 One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to sin(θ) = − 1 is θ = − π . Hence, the family of Quadrant IV answers to number 2 above could just 2 6 have easily been written θ = − π + 2πk for integers k . While on the surface, this family may look 6 10.2 The Unit Circle: Cosine and Sine different than the stated solution of θ = they represent the same list of angles. 10.2.1 11π 6 625 + 2πk for integers k , we leave it to the reader to show Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In defining the cosine and sine func...
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