Tan3x=3Tanx-Tan^3x by 1-3tan^2x Prove it?

2 Answers
Apr 18, 2018

Kindly go through a Proof in the Explanation.

Explanation:

We have, #tan(x+y)=(tanx+tany)/(1-tanxtany)............(diamond)#.

Letting #x=y=A#, we get,

#tan(A+A)=(tanA+tanA)/(1-tanA*tanA)#.

#:. tan2A=(2tanA)/(1-tan^2A)............(diamond_1)#.

Now, we take, in #(diamond), x=2A, and, y=A#.

#:. tan(2A+A)=(tan2A+tanA)/(1-tan2A*tanA)#.

#:. tan3A={(2tanA)/(1-tan^2A)+tanA}/{1-(2tanA)/(1-tan^2A)*tanA}#,

#={(2tanA+tanA(1-tan^2A))/(1-tan^2A)}-:{1-(2tan^2A)/(1-tan^2A)}#,

#=(2tanA+tanA-tan^3A)/(1-tan^2A-2tan^2A)#.

# rArr tan3A=(3tanA-tan^3A)/(1-3tan^2A)#, as desired!

Apr 18, 2018

Let's do it from first principles from De Moivre:

#cos 3 x + i sin 3x = (cos x + i sin x)^3 #

Using the #1,3,3,1# row of Pascal's triangle,

#cos 3 x + i sin 3x#

# = cos^3 x + 3 \cos^2 x (i \sin x) + 3 \cos x (i^2 \sin^2 x ) + i^3 sin^3 x #

# = (cos^3 x- 3 \cos x \sin^2 x ) + i ( 3 \cos^2 x \sin x - sin^3 x)#

Equating respective real and imaginary parts,

# \cos 3 x = cos^3 x- 3 \cos x \sin^2 x#

#\sin 3x = 3 \cos ^2 x \sin x - \sin ^3 x#

Those are (a fairly obscure form of) the triple angle formulas, and typically we'd just write those or a more standard form down and start from here.

# \tan 3x = \frac{ sin 3x}{cos 3x } = frac{ 3 \cos ^2 x \sin x - \sin ^3 x}{ cos^3 x- 3 \cos x \sin^2 x} cdot \frac{1/cos^3 x}{1/ cos^3 x}#

#tan 3x = \frac{3 tan x - tan^3 x}{1 - 3 tan ^2 x} \quad \square#