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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
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Have you tried using\[\tan(3x)=\tan(2x+x)\] Then \[\tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\]
the question asked to solve it using tan3x = sin3x / cos3x :/
|dw:1369139564173:dw|

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Other answers:

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

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