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anonymous

  • 4 years ago

f(θ) = (3cosθ−sinθ)^2

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  1. anonymous
    • 4 years ago
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    simplify the espression

  2. Mertsj
    • 4 years ago
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    \[(3\cos \theta-\sin \theta)^2=9\cos ^2\theta-6\sin \theta \cos \theta+\sin ^2\theta\]

  3. Mertsj
    • 4 years ago
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    Remember that sin^2(x) + cos^2(x)=1

  4. anonymous
    • 4 years ago
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    1. f(θ) = 5−3sin2θ+4cos2θ 2. f(θ) = 4−3sin2θ+5cos2θ 3. f(θ) =−3+5sin2θ+4cos2θ 4. f(θ) = 4+4sin2θ−3cos2θ 5. f (θ) = 5+4sin2θ−3cos2θ 6. f(θ) = 3+4sin2θ+5cos2θ

  5. anonymous
    • 4 years ago
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    its one of those answers

  6. anonymous
    • 4 years ago
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    but idk which

  7. anonymous
    • 4 years ago
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    6sinθcosθ = 3sin2θ - thats one bit of it done

  8. Mertsj
    • 4 years ago
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    1st one

  9. anonymous
    • 4 years ago
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    ahh yes 8 cos^2 θ + cos^2θ + sin^2θ = 8cos^2θ + 1 and 8 cos^2 θ = 4cos2θ + 4 combine to give 1st one

  10. Mertsj
    • 4 years ago
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    \[9\cos^2\theta=9(1-\sin^2\theta)=9-9\sin^2\theta\] Now add the sin^2 theta \[9-9\sin^2\theta+\sin^2\theta=9+8\sin^2\theta\]

  11. Mertsj
    • 4 years ago
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    But \[\sin^2\theta=\frac{1-\cos2\theta}{2}\]

  12. Mertsj
    • 4 years ago
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    Whoops. Sign error 9-8sin^2 theta

  13. anonymous
    • 4 years ago
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    yup - very nice

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