trig

[vanalm]

I'm taking trig online and the book does not cover some of the problems we are supposed to do. I am to verify the identity.

tan8k-tan8k tan^2 4k= 2 tan 4k

Our book briefly explains that tan2x = tan (x+x). So I assume that tan8k= tan(k+k+k+k+k+k+k+k). As far as factoring goes, would I pull out a 4k? or a tan4k? Or can I not pull anything out? The book does not tell what you are allowed to pull out, or if you are even allowed to factor. I'm a little confused on what steps to take with this problem. Any help would be appreciated. Thanks.

 

[royhaas]

Since \(\displaystyle tan(8x)=tan(4x+4x)\), start with the formula for the tangent of twice an angle.

 

[soroban]

Hello, vanalm!

\(\displaystyle \tan8k\,-\,\tan8k\cdot\tan^24k\;=\;2\cdot\tan4k\)

We have the formula: \(\displaystyle \,\tan2\theta\;=\;\frac{2\cdot\tan\theta}{1\,-\,\tan^2\theta}\)

In particular: \(\displaystyle \:\tan8k\;=\;\frac{2\cdot\tan4k}{1\,-\,tan^24k}\)


Factor the left side: \(\displaystyle \,\tan8k\,-\,\tan8k\cdot\tan^24k \;= \;\tan8k\cdot(1\,-\,\tan^24k)\)

Using the formula, we have: \(\displaystyle \,\frac{2\cdot\tan4k}{1\,-\,\tan^24k}\cdot(1\,-\,\tan^24k) \;= \;2\cdot\tan4k\)