**Unformatted text preview: **as usual, this is something we
need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to ﬁnish
the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2 (θ) + sin2 (θ) = 1,
√
to get cos2 (θ) + x2 = 1, or cos(θ) = ± 1 − x2 . Since − π ≤ θ ≤ π , cos(θ) ≥ 0, and thus
2
2
√
√
cos(θ) = 1 − x2 . Our ﬁnal answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2 .
3. We start with the right hand side of the identity and note that 1 + tan2 (θ) = sec2 (θ). From
this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms
of cos(θ) and sin(θ):
2 tan(θ)
1 + tan2 (θ) = 2 tan(θ)
sec2 (θ)
2 = sin(θ)
cos(θ)
1
cos2 (θ) =2 sin(θ)
cos(θ) =2 sin(θ)
cos($
$$θ ) cos(θ )
cos($
$$θ ) cos2 (θ) = 2 sin(θ) cos(θ)
= sin(2θ)
4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos2 (θ) − 1, expresses
cos(2θ) as a polynomial in terms of cos(θ). We are now asked to ﬁnd such an identity for
cos(3θ). Using the sum formula for cosine, we begin with
cos(3θ) = cos(2θ + θ)
= cos(2θ) cos(θ) − sin(2θ) sin(θ)
Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute
cos(2θ) = 2 cos2 (θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields
cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ)
= 2 cos2 (θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ)
= 2 cos3 (θ) − cos(θ) − 2 sin2 (θ) cos(θ) 10.4 T...

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