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anonymous

  • one year ago

1+cos(18x)=_____. a) 2sin^2 (18x) b) 2cos^2 (9x) c) 2cos^2 (18x) d) 2sin^2 (9x)

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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. anonymous
    • one year ago
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    @mathstudent55

  3. anonymous
    • one year ago
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    hi need help

  4. anonymous
    • one year ago
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    yes i need help

  5. anonymous
    • one year ago
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    i gave you a medal

  6. anonymous
    • one year ago
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    with what

  7. jim_thompson5910
    • one year ago
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    \[\Large 1 + \cos(18x) = 1 + \cos(2*9x)\] \[\Large 1 + \cos(18x) = 1 + 2\cos^2(9x) - 1\] \[\Large 1 + \cos(18x) = 2\cos^2(9x)\] On step 2, I used this identity \[\Large \cos(2x) = \cos^2(x) - 1\] see this link for more trig identities http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf specifically see page 2 and the "Double Angle Formulas" section

  8. jim_thompson5910
    • one year ago
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    oops I made a typo, I meant to say \[\Large \cos(2x) = 2\cos^2(x) - 1\]

  9. anonymous
    • one year ago
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    how do i use that equation to find the answer

  10. anonymous
    • one year ago
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    @jim_thompson5910

  11. anonymous
    • one year ago
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    someone please help

  12. jim_thompson5910
    • one year ago
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    I practically gave you the answer. Read through my steps again.

  13. anonymous
    • one year ago
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    oh duh.

  14. anonymous
    • one year ago
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    haha

  15. anonymous
    • one year ago
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    thanks!

  16. jim_thompson5910
    • one year ago
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    You use the identity to replace cos(2*9x) with 2*cos^2(9x) - 1 the "9x" is thought of as the theta in \[\Large \cos(2\theta) = 2\cos^2(\theta) - 1\]

  17. jim_thompson5910
    • one year ago
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    np

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