# Show that secx + rt3 cosecx = 4

1. Dec 19, 2008

### matt_crouch

1. The problem statement, all variables and given/known data

Show that secx + rt3 cosecx = 4 can be written in the form sinx + rt3cos x = 2sin2x

2. Relevant equations

3. The attempt at a solution

Didnt really know where to start with this one. trig identities maybe?
cheers

2. Dec 19, 2008

### tiny-tim

Hi matt_crouch!
oh come on!

whenever you see sec or cosec, multiply everything by cos and/or sin

3. Dec 19, 2008

### matt_crouch

Re: Trig

Im probably missing a really obvious step. But i still dont know where to start.

If i write the equation as 1/cosx + 1/rt3sinx = 4 and i multiply by sinx and cosx to get

1 + 1/rt3 =4cosxsinx

and im not sure but can the RHS be written as 2sin2x?
am i heading in the right direction cos im still stuck as to how i can get 1+ 1/rt3 can be written as sinx + rt3cosx

4. Dec 19, 2008

### Dick

Re: Trig

That's some pretty nasty algebra on the RHS. You are multiplying by sin(x)*cos(x). How did you get the 1's and why did sqrt(3) move into the denominator?

5. Dec 19, 2008

### tiny-tim

erm … sinx + cosx/rt3 =4cosxsinx

but how did that rt3 get on the bottom?
Yes!!!

You must learn these trigonometric identities and be sure of them!

6. Dec 20, 2008

### matt_crouch

Re: Trig

ya i really have.. =]

basically i had no idea what i was doing. Confused the hell out of myself so tore the page out an started again.. =]

7. Dec 20, 2008

### Дьявол

Re: Trig

Sorry for asking, but what is rt3?

8. Dec 20, 2008

### tiny-tim

i assumed it was √3 … but does it matter?

9. Dec 21, 2008

### Дьявол

Re: Trig

In this case, it doesn't. I wanted to know in case if I find rt3 again somewhere. Thanks tiny-tim.