anonymous
  • anonymous
Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π) : tan θ + cot θ = 4 sin 2θ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Callisto
  • Callisto
Hmm... \[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]
anonymous
  • anonymous
I did that and after that I got: \[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]
anonymous
  • anonymous
after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]

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More answers

Callisto
  • Callisto
2cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?
anonymous
  • anonymous
\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]
Callisto
  • Callisto
Multiply both sides by sin2θ, what do you get?
anonymous
  • anonymous
wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]
Callisto
  • Callisto
The first one.
anonymous
  • anonymous
wait, haha I'm sorry!...
anonymous
  • anonymous
its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]
Callisto
  • Callisto
Next: Use cos2x = 1-2(sinx)^2
anonymous
  • anonymous
so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]
anonymous
  • anonymous
riiiight! I see it.
Callisto
  • Callisto
\[cos2x = 1-2sin^2x\]\[2sin^2x = 1-cos2x\]Replace x by 2θ. Life isn't difficult then.
Callisto
  • Callisto
It should be\[\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta\]
anonymous
  • anonymous
i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]
Callisto
  • Callisto
Are you sure...???
Callisto
  • Callisto
\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]
anonymous
  • anonymous
2theta =pi/4, 3pi/4
anonymous
  • anonymous
so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??
anonymous
  • anonymous
yupp
anonymous
  • anonymous
now, how do I find the the equation in the interval [0, 2π)??
Callisto
  • 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 \[sin2\theta = -\sqrt{\frac{1}{2}}\]
anonymous
  • anonymous
i noticed something, since the square root was taken to take out the squared sine....then the result needs to be +- \[\pm \frac{ \sqrt{2} }{ 2 }\]
Callisto
  • Callisto
:)
anonymous
  • anonymous
wow! at the same time
Callisto
  • 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.
anonymous
  • anonymous
ohhhh ok
Callisto
  • Callisto
But it's up to you :)
anonymous
  • anonymous
ok, so when I solve for all the solutions...is it?: \[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]
anonymous
  • anonymous
From zero to pi, yes... keep going to 2pi... 4 more solutions.
anonymous
  • anonymous
The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.
Callisto
  • 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)
Callisto
  • Callisto
What have you put?
anonymous
  • anonymous
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
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
I don't like to just give answers, but you are working very hard at this. ;)
Callisto
  • 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 :(
anonymous
  • anonymous
My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.
anonymous
  • anonymous
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.
anonymous
  • anonymous
its [0, 2pi)
anonymous
  • anonymous
ok... I stand by my answer... and @Callisto 's
anonymous
  • anonymous
ok. hahahah thank you so sooooo much! and I'm sorry for the trouble
anonymous
  • anonymous
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.
anonymous
  • anonymous
ok. thanks!
Callisto
  • Callisto
Agree with @EulerGroupie 's saying :)

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