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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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