Stitz-Zeager_College_Algebra_e-book

# For the scalar property assume that v v1 v2 and w w1

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Unformatted text preview: y 2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that 7 cot(2θ) = 24 . Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to ﬁnd cos(θ) and sin(θ), which means we have two options. If we use the arccotangent function immediately, after the usual calculations we 7 get θ = 1 arccot 24 . To get cos(θ) and sin(θ) from this, we would need to use half angle 2 7 identities. Alternatively, we can start with cot(2θ) = 24 , use a double angle identity, and 7 then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 24 , we have 2 tan(2θ) = 24 . Using the double angle identity for tangent, we have 1−tan(θ) ) = 24 , which 7 7 tan2 (θ 2 (θ ) + 14 tan(θ ) − 24 = 0. Factoring, we get 2(3 tan(θ ) + 4)(4 tan(θ ) − 3) = 0 which gives 24 tan 4 gives tan(θ) = − 3 or tan(θ) = 3 . While either of these values of tan(θ) satisﬁes the equation 4 7 cot(2θ) = 24 , we choose tan(θ) = 3 ,...
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