Half Angles

[vanbeersj]

In studying interference patterns of radio signals, the expression 2E^2-2E^2cos(pie-theta) arises. Show that this can be written as 4E^2cos^2(theta/2).

Sorry I don't know how to put the symbols for theta and pie in.

I started this problem out by factoring out the 2E^2
I get 1-cos(pie)+cos(theta), already I think i started wrong. I need help starting the problem, I'm not completely familiar with half angles, they aren’t explained well in my txt.

Thanks.

 

[Deleted member 4993]

In studying interference patterns of radio signals, the expression 2E^2-2E^2cos(pie-theta) arises. Show that this can be written as 4E^2cos^2(theta/2).

Sorry I don't know how to put the symbols for theta and pie in.

I started this problem out by factoring out the 2E^2
I get 1-cos(pie)+cos(theta), <<< How's that? Cos(A-B) = Cos(A)*Cos(B) - Sin(A)*Sin(B)

also you need to use:

Cos(2?) = Cos[sup9org7z7]2[/sup9org7z7](?) - Sin[sup9org7z7]2[/sup9org7z7](?) = 2Cos[sup9org7z7]2[/sup9org7z7](?) - 1



already I think i started wrong. I need help starting the problem, I'm not completely familiar with half angles, they aren’t explained well in my txt.

Thanks.

 

[chrisr]

Cos(pi-theta) is not a product of two factors in the way that x(a-b) is xa-xb.
Cos(A-B) is the Cosine of the angle formed by subtracting B from A.
Cos(30-20) is Cos10, not Cos30-Cos20.

To solve, you use the techniques given in the response to your post.

Be clear you know what Sine, Cosine and Tangent refer to first.

When you suspected you started out wrongly, you were right!

 

[soroban]

\(\displaystyle \text{Hello, vanbeersj!}\)

\(\displaystyle \text{We need two identities: }\;\begin{array}{ccc}\cos(\pi-\theta) \:=\: -\cos\theta \\ \\[-3mm]\dfrac{1+\cos\theta}{2} \:=\: \cos^2\!\left(\frac{\theta}{2}\right) \end{array}\)

\(\displaystyle \text{Show that }\,2e^2 - 2e^2\!\cos(\pi-\theta)\text{ can be written as: }\:4e^2\!\cos^2\!\left(\tfrac{\theta}{2}\right)\)

\(\displaystyle \text{We have: }\;2e^2-2e^2\!\underbrace{\cos(\pi - \theta)}_{\text{This is }-\cos\theta} \;=\;2e^2 + 2e^2\!\cos\theta\)

. . . . . \(\displaystyle =\; 2e^2(1 + \cos\theta) \;=\;4e^2\underbrace{\left(\frac{1+\cos\theta}{2}\right)}_{\text{This is }\cos^2\!\frac{\theta}{2}} \;=\;4e^2\!\cos^2\!\left(\tfrac{\theta}{2}\right)\)

 

[vanbeersj]

My txt didn't provide me with the first identity. Thank You for all your help.

 

[Deleted member 4993]

You did know:

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

and cos(?) = -1 and sin(?) = 0.

The first identity can be derived from above.