Similarly for sin(θ/2).Proving Trig Identities Using Complex Exponentials

[fredrick08]

Homework Statement

use the definitions of the trig functions in terms of complex exponential to prove:
cos(\theta/2)=\pmsqrt(1+cos(\theta)/2) and
sin(\theta/2)=\pmsqrt(1-cos(\theta)/2)

Homework 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)

The Attempt at a Solution

Ok I am 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 I am unsure how the question wants to be done? it says use complex exponetials?? please can anyone put me on the right track?

 

[tiny-tim]

Hi fredrick08!

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

[use the definitions of the trig functions in terms of complex exponential to prove:
cos(\theta/2)=\pmsqrt(1+cos(\theta)/2) and
sin(\theta/2)=\pmsqrt(1-cos(\theta)/2)

Your equations are wrong.

They should be:

cos(θ/2) = ±√((1 + cosθ)/2)

sin(θ/2) = ±√((1 - cosθ)/2)​

Does that help?

 

[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 don't understand how to attempt the problem...

 

[tiny-tim]

oops!

sorry! i put the brackets in the wrong place.

I've edited and corrected it now.

(and try e^iθ = (e^iθ/2)² )

 

[fredrick08]

ahh thx

 

[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?

 

[tiny-tim]

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

 

[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?

 

[tiny-tim]

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

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

by equating the imaginary parts, you've just proved it.

… 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?

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? ​

 

[fredrick08]

ah i see, thanks heaps

 

[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

 

[fredrick08]

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

 

[tiny-tim]

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

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

(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).