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



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

AonZ
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the question asked to solve it using tan3x = sin3x / cos3x :/

AonZ
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dw:1369139564173:dw

AonZ
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can you pls post it :)
May help me solve it :)

trumpetmaster7777
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factor out sin and cos

trumpetmaster7777
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(3sinx  4sin^3 x)/(4cos^3 x  3cosx)
(sinx)(34sin^2(x))/((cosx)(4cos^2(x)3)

AonZ
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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

trumpetmaster7777
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Pick a trig function, sine or cosine

AonZ
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any :)

trumpetmaster7777
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Thanks, loser66

trumpetmaster7777
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so, let's go with sine
cos^2(x)=1sin^2(x), right?

AonZ
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yep

trumpetmaster7777
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plug in (1sin^2(x)) for cos^2(x)
(sinx)(34sin^2(x))/((cosx)(4cos^2(x)3)

AonZ
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yea i actually tried that :P

AonZ
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after doing that would you divide by cos^2 x?

trumpetmaster7777
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wait up
look at the tan(3x) expression
can't you clean that up a bit?

AonZ
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possibly... But you could always change it both ways

AonZ
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this is the last step i got todw:1369140948450:dw

trumpetmaster7777
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1/cos^2(x) is sec^2(x), right?

trumpetmaster7777
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use the tangentsecant identity

trumpetmaster7777
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1+tan^2(x) = sec^2(x)

AonZ
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!!!! :D ur awsome

trumpetmaster7777
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Thank you :D

AonZ
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on denominator its still sec^2  4tan^2 ... errr
im lost again

RolyPoly
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Would that help?
sec^2 = 1 + tan^2

AonZ
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oh yes :)