## anonymous 3 years ago Find the exact value by using a half-angle identity. tan7π/8 PLEASE PLEASE PLEASE :)

1. anonymous

Well, the half angle identity says: $\sin(\theta/2) = \sqrt{\frac{1-\cos(\theta)}{2}} \quad \cos(\theta/2) = \sqrt{\frac{1+\cos(\theta)}{2}}$

2. anonymous

3. 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.

4. anonymous

So then $\tan(\theta / 2) = \sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}$

5. anonymous

well i just don't really understand how to do the problem once everything is plugged in

6. anonymous

Well, $$\cos(7\pi/4)$$ is relatively easy to solve for, when you look at the unit circle.

7. anonymous

well i know how to find the degree but once i get the degree i dont know where to go from there

8. anonymous

that is a good method... but tan's identity is below $\large \tan (\frac{ \theta }{ 2 }) =\frac{ \sin(\theta) }{1+ \cos (\theta) }=\frac{ 1- \cos (\theta) }{\sin(\theta) }$

9. anonymous

do you mind just showing me the steps please??

10. anonymous

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11. anonymous

can you use the half angle formula

12. anonymous

yes, if you consider 7pi/8 to be a half angle, then that means the full angle is 7pi/4

13. anonymous

Since for 7pi/4 it is easy to find the values of sin/cos, it is useful here.

14. 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.

15. 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

16. anonymous

$$a^2+a^2=1$$ is just Pythagorean theorem.

17. anonymous