anonymous
  • anonymous
Find the exact value by using a half-angle identity. tan7π/8 PLEASE PLEASE PLEASE :)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Well, the half angle identity says: \[ \sin(\theta/2) = \sqrt{\frac{1-\cos(\theta)}{2}} \quad \cos(\theta/2) = \sqrt{\frac{1+\cos(\theta)}{2}} \]
anonymous
  • anonymous
but what about tan..
anonymous
  • anonymous
Well \[ \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} \]So I didn't really remember it's identity for this. Rather, I just divided the other identities.

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anonymous
  • anonymous
So then \[ \tan(\theta / 2) = \sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}} \]
anonymous
  • anonymous
well i just don't really understand how to do the problem once everything is plugged in
anonymous
  • anonymous
Well, \(\cos(7\pi/4)\) is relatively easy to solve for, when you look at the unit circle.
anonymous
  • anonymous
well i know how to find the degree but once i get the degree i dont know where to go from there
anonymous
  • anonymous
that is a good method... but tan's identity is below \[\large \tan (\frac{ \theta }{ 2 }) =\frac{ \sin(\theta) }{1+ \cos (\theta) }=\frac{ 1- \cos (\theta) }{\sin(\theta) }\]
anonymous
  • anonymous
do you mind just showing me the steps please??
anonymous
  • anonymous
|dw:1353034351647:dw|
anonymous
  • anonymous
can you use the half angle formula
anonymous
  • anonymous
yes, if you consider 7pi/8 to be a half angle, then that means the full angle is 7pi/4
anonymous
  • anonymous
Since for 7pi/4 it is easy to find the values of sin/cos, it is useful here.
anonymous
  • anonymous
Think of it this way, 7pi/4 is pi/4 short of 2pi. 2pi is a full circle. So pi/4 is just a quarter of a circle.
anonymous
  • anonymous
What makes a quarter of a circle easy to calculate, is the fact that the you know the the horizontal and vertical part are equal, so I set them to a and solved for a to find the value of cos
anonymous
  • anonymous
\(a^2+a^2=1\) is just Pythagorean theorem.
anonymous
  • anonymous
https://www.youtube.com/watch?v=Nh4Ro86dnHk

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