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**Unformatted text preview: **. The latter form
3
of the solution is best understood looking at the geometry of the situation in the diagram
below.3
3 See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simpliﬁcation of the solution. 10.3 The Six Circular Functions and Fundamental Identities
y
1 639
y
1 π
3
x
1 x
1 π
3 3. From the table of common values, we see that π has a cotangent of 1, which means the
4
solutions to cot(θ) = −1 have a reference angle of π . To ﬁnd the quadrants in which our
4
solutions lie, we note that cot(θ) = x , for a point (x, y ), y = 0, on the Unit Circle. If cot(θ) is
y
negative, then x and y must have diﬀerent signs (i.e., one positive and one negative.) Hence,
π
our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 34 + 2πk , and for
π
Quadrant IV, we get θ = 74 + 2πk for integers k . Can these lists be combined? We see that,
π
in fact, they can. One way to capture all the solutions is: θ = 34 + πk for integers k .
y
1 y
1 π
4
x
1 π
4 x
1 We have already seen the importan...

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