Stitz-Zeager_College_Algebra_e-book

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ,...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online