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**Unformatted text preview: **3θ) as a polynomial in terms of cos(θ).
Solution.
1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we ﬁnd r = x2 + y 2 = 5.
3
Hence, cos(θ) = − 5 and sin(θ) = 4 . Applying Theorem 10.17, we get cos(2θ) = cos2 (θ) −
5
2
2
7
4
3
sin2 (θ) = − 3 − 4 = − 25 , and sin(2θ) = 2 sin(θ) cos(θ) = 2 5 − 5 = − 24 . Since both
5
5
25
cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position,
lies in Quadrant III. 662 Foundations of Trigonometry 2. If your ﬁrst reaction to ‘sin(θ) = x’ is ‘No it’s not, cos(θ) = x!’ then you have indeed learned
something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a
variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle
which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x
represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms
of x. We will see more of this kind of thing in Section 10.6, and,...

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