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clara1223

  • one year ago

simplify the following, where k represents any integer: cos(2kpi-x)(-sin(2kpi-x)) a) -cosx*sinx b) (1/2)sin(2x) c) cos((1/2)x)^2 d) cos(x)^2*sin(x)^2 e) undefined

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  1. clara1223
    • one year ago
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    @Loser66 could you explain how you got that answer please?

  2. mathstudent55
    • one year ago
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    Here is the explanation: Since the sine and cosine are periodic functions with a period of 2pi, sin (2kpi - x) = sin (-x) and cos(2kpi - x) = cos (-x) Ok so far?

  3. mathstudent55
    • one year ago
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    Then because of even and odd functions, sin (-x) = - sin x and cos (-x) = cos x

  4. mathstudent55
    • one year ago
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    Putting it all together, you get: \((\cos (2k \pi-x)) ((-\sin(2k \pi-x)) \) \(= \cos (-x) \sin (-x)\) \(= - \sin x \cos x\)

  5. clara1223
    • one year ago
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    @mathstudent55 thanks so much!

  6. dumbcow
    • one year ago
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    wait ...... -sin(-x) = sin(x) it equals positive sin(x) cos(x) then use double angle identity for sin ----> sin(2x) = 2sincos ----> sincos = 1/2 sin(2x)

  7. mathstudent55
    • one year ago
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    @dumbcow You're correct. I missed the negative with the sin in the original problem.

  8. mathstudent55
    • one year ago
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    This is how it should read: \((\cos (2k \pi-x)) ((-\sin(2k \pi-x))\) \(= \cos (-x)(- \sin (-x))\) \(= \cos x (-(-\sin x))\)\) \(= \sin x \cos x\) \(=\dfrac{1}{2} \sin 2x\)

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