Proving the Double-Angle Formula for Tangent with Step-by-Step Solution

[odolwa99]

Having some trouble with this one. Can anyone help me out?

Many thanks.

Homework Statement

Prove that \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan\theta}

Homework Equations

The Attempt at a Solution

\frac{3\sin\theta}{\cos\theta}-\frac{\sin^3\theta}{\cos^3\theta}/1-\frac{3\sin^2\theta}{\cos^2\theta}
\frac{3\sin\theta\cos\theta-\sin^3\theta}{\cos^2\theta-3\sin^2\theta}
\frac{\sin\theta(3\cos\theta-\sin^2\theta)}{\cos^2\theta-3\sin^2\theta}
...

 

[rock.freak667]

It will be easier to use the left side to prove the right side.


tan3θ = tan(2θ+θ)

Expand out tan(2θ+θ), what do you get?

 

[vela]

Homework Statement

Prove that \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan\theta}

The tangent in the denominator should be squared.

The Attempt at a Solution

\frac{3\sin\theta}{\cos\theta}-\frac{\sin^3\theta}{\cos^3\theta}/1-\frac{3\sin^2\theta}{\cos^2\theta}

Use parentheses! That should be
\left( \frac{3\sin\theta}{\cos\theta} - \frac{\sin^3\theta}{\cos^3\theta}\right) / \left(1-\frac{3\sin^2\theta}{\cos^2\theta}\right)

\frac{3\sin\theta\cos\theta-\sin^3\theta}{\cos^2\theta-3\sin^2\theta}

Check your algebra. That line doesn't follow from the one above.

 

[odolwa99]

Ok, thank you. I'll give that a 2nd look.