Thread: Trig proof - how do I prove this?

Prove that

$\displaystyle tan 4 theta = 4 tan theta (1-tan^2 theta)/1-6tan^2 theta + tan ^4 theta$

(Sorry about that. I don't know how to use LaTEX, it's all messed up. I used the MATH tag and it showed an error and so I used the TEX tag).

I don't know where to start. What trig identities should I use?

Online LaTeX Equation Editor - create, integrate and download it's easier.

$\displaystyle \tan(4\theta )=\frac{4\tan(\theta) (1-\tan^{2}(\theta))}{1-6\tan^2(\theta)+\tan^4(\theta)}$

Use twice $\displaystyle \tan(2\theta )=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$.

Originally Posted by Don

Prove that

$\displaystyle \tan 4 \theta = 4 \tan \theta (1-\tan^2 \theta)/(1-6\tan^2 \theta + \tan ^4 \theta)$

(Sorry about that. I don't know how to use LaTEX, it's all messed up. I used the MATH tag and it showed an error and so I used the TEX tag).

I don't know where to start. What trig identities should I use?

(Use the TEX tags, but put a backslash before each tan and each theta.)

Use the formula $\displaystyle \tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}.$ Then use the same formula again to get $\displaystyle \tan(4\theta)$ in terms of $\displaystyle \tan(2\theta).$ Finally, combine the two formulas to get $\displaystyle \tan(4\theta)$ in terms of $\displaystyle \tan\theta.$

Got it, thanks.