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Stitz-Zeager_College_Algebra_e-book

# 3 and the answer to exercise 7 in section 64 1110

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Unformatted text preview: w1 = − 2+ i 2, w2 = − 2 − i 2, to √ and w3 = 2 − i 2. 854 Applications of Trigonometry √ √ 3. For z = 2+ i 2, we have z = 2cis π . With r = 2, θ = π and n = 3 the usual computations 4 4 √ √ √ √ π π 7π yield w0 = 3 2cis 12 , w1 = 3 2cis 9π = 3 2cis 34 and w2 = 3 2cis 112 . If we were 12 to convert these to rectangular form, we would need to use either the Sum and Diﬀerence Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w0 and w2 . Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. 4. To ﬁnd the ﬁve ﬁfth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. √ π π π Since 5 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 25 , w2 = cis 45 , w3 = cis 65 and π w4 = cis 85 . The situation here is even graver than in the previous example, since we have π no identities developed to help us determine the cosine or sine of 25 . At this stage, we could approximate our answers using a calculator, and we leave this to the Exercises. Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Es...
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