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AonZ

  • one year ago

Pls Pls Pls help... I have been stuck for a while!!! Prove tan3x = (3tanx - tan^3 x) / ( 1 - 3tan^2 x) using sin3x/ cos3x = tan3x sin3x = 3sinx - 4sin^3 x cos3x = 4cos^3 x - 3cosx

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  1. .Sam.
    • one year ago
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    Have you tried using\[\tan(3x)=\tan(2x+x)\] Then \[\tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\]

  2. AonZ
    • one year ago
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    the question asked to solve it using tan3x = sin3x / cos3x :/

  3. AonZ
    • one year ago
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    |dw:1369139564173:dw|

  4. AonZ
    • one year ago
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    can you pls post it :) May help me solve it :)

  5. trumpetmaster7777
    • one year ago
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    factor out sin and cos

  6. trumpetmaster7777
    • one year ago
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    (3sinx - 4sin^3 x)/(4cos^3 x - 3cosx) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)

  7. AonZ
    • one year ago
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    my way was to factor out sin and cos so it was tanx times something then i divided the numerators by cos^2 x then im stuck

  8. trumpetmaster7777
    • one year ago
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    Pick a trig function, sine or cosine

  9. AonZ
    • one year ago
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    any :)

  10. trumpetmaster7777
    • one year ago
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    Thanks, loser66

  11. trumpetmaster7777
    • one year ago
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    so, let's go with sine cos^2(x)=1-sin^2(x), right?

  12. AonZ
    • one year ago
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    yep

  13. trumpetmaster7777
    • one year ago
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    plug in (1-sin^2(x)) for cos^2(x) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)

  14. AonZ
    • one year ago
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    yea i actually tried that :P

  15. AonZ
    • one year ago
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    after doing that would you divide by cos^2 x?

  16. trumpetmaster7777
    • one year ago
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    wait up look at the tan(3x) expression can't you clean that up a bit?

  17. AonZ
    • one year ago
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    possibly... But you could always change it both ways

  18. AonZ
    • one year ago
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    this is the last step i got to|dw:1369140948450:dw|

  19. trumpetmaster7777
    • one year ago
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    1/cos^2(x) is sec^2(x), right?

  20. trumpetmaster7777
    • one year ago
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    use the tangent-secant identity

  21. trumpetmaster7777
    • one year ago
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    1+tan^2(x) = sec^2(x)

  22. AonZ
    • one year ago
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    !!!! :D ur awsome

  23. trumpetmaster7777
    • one year ago
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    Thank you :D

  24. AonZ
    • one year ago
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    on denominator its still sec^2 - 4tan^2 ... errr im lost again

  25. RolyPoly
    • one year ago
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    Would that help? sec^2 = 1 + tan^2

  26. AonZ
    • one year ago
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    oh yes :)

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