## Dama28 Group Title Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π) : tan θ + cot θ = 4 sin 2θ one year ago one year ago

1. Callisto

Hmm... $tan θ + cot θ$$= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?$

2. Dama28

I did that and after that I got: $\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2$

3. Dama28

after that: $\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta$

4. Callisto

2cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?

5. Dama28

$\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta$

6. Callisto

Multiply both sides by sin2θ, what do you get?

7. Dama28

wait! would it be $1=2\sin ^{2}2\Theta$ or $1= 2\sin^2 4\theta$

8. Callisto

The first one.

9. Dama28

wait, haha I'm sorry!...

10. Dama28

its now $\frac{ 1 }{ 2 }=\sin^2 \theta$

11. Callisto

Next: Use cos2x = 1-2(sinx)^2

12. Dama28

so I take the square root?? and get sin($\sin\frac{ \sqrt{2} }{ 2 }=2\theta$

13. Dama28

riiiight! I see it.

14. Callisto

$cos2x = 1-2sin^2x$$2sin^2x = 1-cos2x$Replace x by 2θ. Life isn't difficult then.

15. Callisto

It should be$\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta$

16. Dama28

i get $2\Theta = \frac{ \pi }{ 2}$ and $\frac{ 3\pi }{ 2 }$

17. Callisto

Are you sure...???

18. Callisto

$sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$

19. hkm91

2theta =pi/4, 3pi/4

20. Dama28

so, $\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }$ ??

21. hkm91

yupp

22. Dama28

now, how do I find the the equation in the interval [0, 2π)??

23. Callisto

Hmm... Something's wrong there. $1=2\sin ^{2}2\theta$$sin^2 2\theta = \frac{1}{2}$$sin2\theta = \pm \sqrt{\frac{1}{2}}$So, $sin2\theta = \sqrt{\frac{1}{2}}$or $sin2\theta = -\sqrt{\frac{1}{2}}$

24. Dama28

i noticed something, since the square root was taken to take out the squared sine....then the result needs to be +- $\pm \frac{ \sqrt{2} }{ 2 }$

25. Callisto

:)

26. Dama28

wow! at the same time

27. Callisto

To avoid taking square root, I prefer using my method: $1=2\sin ^{2}2\theta$$1=1-cos4\theta$$cos4\theta =0$Then. solve theta.

28. Dama28

ohhhh ok

29. Callisto

But it's up to you :)

30. Dama28

ok, so when I solve for all the solutions...is it?: $\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }$

31. EulerGroupie

From zero to pi, yes... keep going to 2pi... 4 more solutions.

32. EulerGroupie

The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.

33. Callisto

If you want to get the next 4 solutions, use: 2pi - solution you've got. Then you'll have 4 more. Eg: 2pi - (pi/8)

34. Callisto

What have you put?

35. Dama28

pi/8, 3pi/8 my first try and pi/8, 3pi/8, 5pi/8, 7pi/8 the second try. I have one more try :S

36. EulerGroupie

pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 are all of the solutions between zero and 2pi. I wish I could show you my pretty graph on my calculator. :)

37. EulerGroupie

I don't like to just give answers, but you are working very hard at this. ;)

38. Callisto

First four: pi/8, 3pi/8, 5pi/8, 7pi/8 Next four: 2pi - pi/8 = 15pi/8 2pi - 3pi/8 = 13pi/8 2pi - 5pi/8 =... 2pi - 7pi/8 =... I'm sorry :(

39. EulerGroupie

My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.

40. Dama28

hahaha thanks. the reason why I didn't keep going is because in the previous problems I had, if it was divided by 3 theta I would only give 6 answers. If if was divided by 4theta, then 8 answers, so I would assume that in this excercise that's divided by 2theta I would only give 4 answers.

41. Dama28

its [0, 2pi)

42. EulerGroupie

ok... I stand by my answer... and @Callisto 's

43. Dama28

ok. hahahah thank you so sooooo much! and I'm sorry for the trouble

44. EulerGroupie

Oh, this stuff is fun... no troubles. We do it because we love it... especially when people are really working and trying instead of fishing for easy answers.

45. Dama28

ok. thanks!

46. Callisto

Agree with @EulerGroupie 's saying :)