# Homework Help: Solving Trig Identity?

1. Dec 8, 2012

### Stanc

1. The problem statement, all variables and given/known data
Need some help finding all solutions for x....

csc^2((x)/(2)) = 2secx

3. The attempt at a solution
Not sure what kind of approach to take but:

1/ sin^2(x/2) = 2/ cos x

From here Not sure what to do i tried cross multiplying and got cos x = 2sin^2(x/2) but got no idea from here.... please help

2. Dec 8, 2012

### Stanc

Does anyone have an explanation for this solution?? someone just sent it:

csc^2(x/2)=2secx
2/(1-cosx)=2secx
2=2secx(1-cosx)
1=secx-1
2=secx
π/3+2πk=x
5π/3+2πk=x

However i do think hes missing a squared? Can anyone explain this formula?

3. Dec 8, 2012

### rock.freak667

cosec2(x/2) = secx

or

½cosec2(x)=secx

If it is the first one, then your double angle formulas for cos2x will help you out here greatly.

4. Dec 9, 2012

### Staff: Mentor

The line above is incorrect on the left side. It should be
2/(1 - cos2(x/2)) = 2sec(x)
To solve this equation, convert everything into terms involving cosine. You need the double angle formula to convert cos(x) into cos(2 * x/2).

BTW, your title is misleading. You do not "solve" an identity - you prove that an equation is identically true (true for all or most values of the variable). What this seems to be is an equation to solve for specific values of x.

5. Dec 9, 2012

### SammyS

Staff Emeritus
I would be inclined to go from where you are here
cos(x) = 2sin^2(x/2)​
and use the double angle formula to convert cos(x) to a form having sin(x/2). In other words, think of cos(x) as cos(2(x/2)) .

Then you will have an expression involving only sin(x/2) .

6. Dec 9, 2012

### Stanc

So basically change cosx into 1-sin^2(x/2) ??

7. Dec 9, 2012

### HallsofIvy

NO, cos x is NOT equal to 1- sin^2(x/2). It is equal to 1- sin^2(x). You could then use the identity sin(x)= 2sin(x/2)cos(x/2).

8. Dec 9, 2012

### Stanc

Sorry for the title too
Sorry i do not follow why you have the 2 over in 2/(1-cos^2(x/2))

9. Dec 9, 2012

### Stanc

Thanks, i wasnt too sure about that part..... But isnt it cos^2x that is suppose to equal 1 - sin^2x?

10. Dec 9, 2012

### ehild

Apply the half-angle formula: sin2(x/2)=(1-cosx)/2

ehild

11. Dec 9, 2012

### Stanc

Yes, i can solve it with the half angle formula but was wondering if i could solve it without that formula....

12. Dec 9, 2012

### ehild

You need to use either the half-angle formula or the double-angle one.

ehild

13. Dec 9, 2012

### Stanc

How would i approach it with double angle? I followed some of the steps and came with this

1/1-cos^2(x/2) = 2/cos2(x/2)

2-2cos^2(x/2) = cos2(x/2)

Is this right?

14. Dec 9, 2012

### SammyS

Staff Emeritus
Not quite:

cos(x) = 1 - 2sin2(x/2)

15. Dec 9, 2012

### Stanc

Oh ya, sorry about that, so from here i have:

1-2sin^2(x/2) = 2sin^2(x/2)

If i move the right side over i would get:

1-4sin^2(x/2) = 0

sin^2(x/2) = 1/4

Square root it:

sin(x/2) = 1/2?

I think thats correct, right?

16. Dec 9, 2012

### ehild

sin(x/2) = ±1/2.

ehild

17. Dec 9, 2012

### Staff: Mentor

That 2 shouldn't be there. The line I was correcting had 2/(1 - cos(x)), and I brought that 2 along, not noticing that it was wrong as well.

18. Dec 9, 2012

### Stanc

Thanks., but the answers i get are pi/6 and 5pi/6 while the answers are pi/3 and 5pi/3

Anything i have to do?

19. Dec 9, 2012

### SammyS

Staff Emeritus
If x/2 = π/6 , then x = π/3 ...