## anonymous one year ago If (2tan(x))/(1-tan^2(x))=1, then what can x equal? My options are: A. x=7pi/8+npi B. x=pi/8+npi C. x=3pi/8+npi D. x=5pi/8+npi

1. anonymous

I think that B is right but I don't know about any others. Can anybody help me?

2. anonymous

I got the question wrong, but I still want to know how to do this!

3. anonymous

$\frac{ 2\tan x }{ 1-\tan ^2x }=1$ $\tan 2x=1=\tan \left( \frac{ \pi }{ 4 }+n \pi \right)$ $x=\left( \frac{ \pi }{ 8 }+n \frac{ \pi }{ 2 } \right)$

4. anonymous

How did you go between the last two steps?

5. anonymous

$\tan 2 x=\frac{ \sin 2x }{ \cos 2x }=\frac{ 2\sin x \cos x }{ \cos ^2x-\sin ^2x }$ divide the numerator and denominator by $\cos ^2x$ $\tan 2x=\frac{ 2\tan x }{ 1-\tan ^2x }$

6. anonymous

$\tan 2x=1=\tan \frac{ \pi }{ 4 }=\tan \left( \frac{ \pi }{ 4 }+n \pi \right)$ $2x=\frac{ \pi }{ 4 }+n \pi,x=?$

7. anonymous

So that's $1/8\pi +1/2n \pi ?$

8. anonymous

correct ,i have written above.

9. anonymous

Okay I get it thank you. That helps a lot.

10. anonymous

You can have a medal :)

11. anonymous

yw

12. anonymous

can you help me with another problem? @surjithayer

13. anonymous

I have $\cos(\pi/4)\cos(\pi/6)=1/2( (blank) \pi/12+\cos(5\pi/12)$

14. anonymous

$2 \cos A \cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ $\cos A \cos B=\frac{ 1 }{ 2 }\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$