AonZ
  • AonZ
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
Have you tried using\[\tan(3x)=\tan(2x+x)\] Then \[\tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\]
AonZ
  • AonZ
the question asked to solve it using tan3x = sin3x / cos3x :/
AonZ
  • AonZ
|dw:1369139564173:dw|

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More answers

AonZ
  • AonZ
can you pls post it :) May help me solve it :)
anonymous
  • anonymous
factor out sin and cos
anonymous
  • anonymous
(3sinx - 4sin^3 x)/(4cos^3 x - 3cosx) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)
AonZ
  • AonZ
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
anonymous
  • anonymous
Pick a trig function, sine or cosine
AonZ
  • AonZ
any :)
anonymous
  • anonymous
Thanks, loser66
anonymous
  • anonymous
so, let's go with sine cos^2(x)=1-sin^2(x), right?
AonZ
  • AonZ
yep
anonymous
  • anonymous
plug in (1-sin^2(x)) for cos^2(x) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)
AonZ
  • AonZ
yea i actually tried that :P
AonZ
  • AonZ
after doing that would you divide by cos^2 x?
anonymous
  • anonymous
wait up look at the tan(3x) expression can't you clean that up a bit?
AonZ
  • AonZ
possibly... But you could always change it both ways
AonZ
  • AonZ
this is the last step i got to|dw:1369140948450:dw|
anonymous
  • anonymous
1/cos^2(x) is sec^2(x), right?
anonymous
  • anonymous
use the tangent-secant identity
anonymous
  • anonymous
1+tan^2(x) = sec^2(x)
AonZ
  • AonZ
!!!! :D ur awsome
anonymous
  • anonymous
Thank you :D
AonZ
  • AonZ
on denominator its still sec^2 - 4tan^2 ... errr im lost again
anonymous
  • anonymous
Would that help? sec^2 = 1 + tan^2
AonZ
  • AonZ
oh yes :)

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