More complex numbers, more trigs

  • #1
rock.freak667
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Homework Statement



The angles [itex]\theta[/itex] and [itex]\phi[/itex] lie on the interval (-pi/2,pi,2) and

[itex]z=(cos\theta +cos\phi)+i(sin\theta +sin\phi)[/itex]

Show that |z|=2cos([itex]\frac{\theta - \phi}{2}[/itex]) and find arg(z)

Homework Equations


If z=x+iy

[tex]|z|=\sqrt{x^2 +y^2}[/tex]


The Attempt at a Solution



[tex]|z|= \sqrt{(cos\theta +cos\phi)^2+(sin\theta +sin\phi)^2}[/tex]

[tex]=2+2cos(\theta - \phi)[/tex]

I don't know how to get rid of the additional two and how to make the angle halved.
 

Answers and Replies

  • #2
rock.freak667
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Solved the first part


2nd part giving trouble


arg(z)=tan-1{Im(z)/Re(z)}

[tex]tan^{-1}(\frac{sin\theta + sin\phi}{cos\theta + cos\phi})[/tex]

I think I'll need to simplify this further.

I tried multiplying by the conjugate of the denominator but that didn't lead anywhere feasible.
 
Last edited:
  • #3
Dick
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(sin(a)+sin(b))/(cos(a)+cos(b))=tan((a+b)/2). I'm going to the wikipedia page of trig identities and fishing around. You might notice this is related to the third problem of your other post if you put phi=0.
 
  • #4
rock.freak667
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(sin(a)+sin(b))/(cos(a)+cos(b))=tan((a+b)/2). I'm going to the wikipedia page of trig identities and fishing around. You might notice this is related to the third problem of your other post if you put phi=0.

oh my god! I completely forgot factor formula....seems that 1 month away from math has completely worked my brain in the wrong way.... thank you.
 
  • #5
Dick
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I don't remember stuff like that either. I just figured that if the problem called for it then something like that must exist.
 
  • #6
rock.freak667
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I don't remember stuff like that either. I just figured that if the problem called for it then something like that must exist.

Well I knew it had to be arctan of the tan of something, but I thought the something would have been just with theta and phi and not half angled. Was looking to make a tan(A+B) thing.
 
  • #7
Dick
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Well I knew it had to be arctan of the tan of something, but I thought the something would have been just with theta and phi and not half angled. Was looking to make a tan(A+B) thing.

See, they tricked you. Seriously, some trig identities are just tricks. It's hard to see them coming. Like that one. If the problem were even slightly different it wouldn't work. Very specific solution to a very specific problem. I think the other problems in your other set are more instructive.
 

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