This preview shows page 1. Sign up to view the full content.
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
2 While, technically speaking, π isn’t a reference angle we can nonetheless use it to ﬁnd our
answers. If we follow the procedure set forth in the previous examples, we ﬁnd θ = π + 2πk
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?)
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
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...
View Full Document