# Trigs and complex exponetials

1. Mar 9, 2010

### fredrick08

1. The problem statement, all variables and given/known data
use the definitions of the trig functions in terms of complex exponential to prove:
cos($$\theta$$/2)=$$\pm$$sqrt(1+cos($$\theta$$)/2) and
sin($$\theta$$/2)=$$\pm$$sqrt(1-cos($$\theta$$)/2)

2. Relevant equations
e^i$$\theta$$=cos$$\theta$$+isin$$\theta$$
e^-i$$\theta$$=cos$$\theta$$-isin$$\theta$$
cos$$\theta$$=1/2(e^i$$\theta$$+e^-i$$\theta$$)
sin$$\theta$$=1/2i(e^i$$\theta$$+e^-i$$\theta$$)

3. The attempt at a solution
Ok im just not sure where to start with this one... my answer obviously has a sqrt in it, and also cos$$\theta$$... do i need cos^2$$\theta$$+sin^2$$\theta$$=1?
or sin2$$\theta$$=2sin$$\theta$$*cos$$\theta$$?

But im unsure how the question wants to be done? it says use complex exponetials?? plz can anyone put me on the right track?

2. Mar 9, 2010

### tiny-tim

Hi fredrick08!

(have a theta: θ and a square-root: √ and a ± )

They should be:

cos(θ/2) = ±√((1 + cosθ)/2)

sin(θ/2) = ±√((1 - cosθ)/2)​

Does that help?

Last edited: Mar 9, 2010
3. Mar 9, 2010

### fredrick08

ok can i ask, On my sheet is says the divide by 2, is in the the sqrt?? are ou saying that is wrong?? Will it make a difference... also with this little difference, i still dont understand how to attempt the problem...

4. Mar 9, 2010

### tiny-tim

oops!

sorry! i put the brackets in the wrong place.

I've edited and corrected it now.

(and try e = (eiθ/2)2 )

5. Mar 9, 2010

ahh thx

6. Mar 9, 2010

### fredrick08

ok i tried what you said.. but i get this awful equation...

cos^2(θ/2)=cos(θ)+i*sin(θ)+sin^2(θ/2)-2i*sin(θ/2)cos(θ/2)....i know this is right... but how, or what do i have to do to it to get the nice equation that i need?

7. Mar 9, 2010

### tiny-tim

Whenever you have a complex equation, you can equate the real parts and the imaginary parts separately.

8. Mar 9, 2010

### fredrick08

ok what i did was change the -2i*sin(θ/2)cos(θ/2) to -i*sin(θ), since they are equal form the a trig identity.. that let mme cancel the imaginary part, so i was left with cos(θ)+sin^2(θ/2), then using another identity that changed to cos(θ)+(1-cos^2(θ/2)) and i know that equals 1/2+cos(θ)/2.... but how do i change my last line into this one???

9. Mar 9, 2010

### tiny-tim

erm … the trig identity, sinθ = 2sin(θ/2)cos(θ/2), comes from this …

by equating the imaginary parts, you've just proved it.
You've got very confused …

the real part of your equation is cos2(θ/2) = cosθ + sin2(θ/2),

and from that you've now got cos2(θ/2) = cosθ + (1 - cos2(θ/2)) …

ok?

10. Mar 9, 2010

### fredrick08

ah i see, thanks heaps

11. Mar 9, 2010

### fredrick08

oh no wait... from cos^2(θ/2) = cosθ + (1 - cos^2(θ/2)) how does this, get to my final answer, what do i have to do to it to get cos^2(θ/2)=1/2+cos(θ)/2

12. Mar 9, 2010

### fredrick08

never mind, double angle formula fixes it, thanks so much for ur help = )

13. Mar 9, 2010

### tiny-tim

(try using the X2 tag just above the Reply box )

No, the double angle formula is what you're trying to prove!

cos2(θ/2) = cosθ + (1 - cos2(θ/2))

So 2cos2(θ/2) = cosθ + 1

So cos(θ/2) = √((cosθ + 1)/2).