Sine, cosine and tangency of angle 11pi/12

  • Thread starter TonyC
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  • #1
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Main Question or Discussion Point

When I worked this problem, I came up with the following:

sin 11pi/12= -sq root 2/4 (sq root3-1)
cos 11pi/12= -sq root 2/4 (sq root3+1)
tan 11pi/12= 2-sq root3

am I far off?
 

Answers and Replies

  • #2
TD
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How did you get that? Does't seem right to me.
 
  • #3
Tide
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How did you arrive at those?

For the first one i get

[tex]sin \frac {11 \pi}{12} = \frac {\sqrt 2}{4} (\sqrt 3 - 1)[/tex]
 
  • #4
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I started by working in radians.
 
  • #5
Tide
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And then what?
 
  • #6
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11pi/12=165 degrees

11pi/12=pi/4+2pi/3
sq root 3/3+sq root 2/2

If sin 11pi/12=sq root 2/4(sq root 3-1)

I came up with:
cos 11pi/12=sq root 2/4(sq root 3+1)
tan 11pi/12=2-sq root3
 
  • #7
TD
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TonyC said:
11pi/12=165 degrees

11pi/12=pi/4+2pi/3
sq root 3/3+sq root 2/2
This is correct: [tex]\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}[/tex]

But [itex]\sin \left( {\alpha + \beta } \right) \ne \sin \left( \alpha \right) + \sin \left( \beta \right)[/tex] so you can't just take the sine of both angles!

Use [itex]\sin \left( {\alpha + \beta } \right) = \sin \left( \alpha \right) \cdot \cos \left( \beta \right) + \cos \left( \alpha \right) \cdot \sin \left( \beta \right)[/itex]
 
  • #8
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I am still confused.....
 
  • #9
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P.S. what program are you using so I don't have to keep writing out the equations long hand?
 
  • #10
TD
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Well, we have that [tex]\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}[/tex]

And we know that:
[tex]\sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}[/tex]
[tex]\sin \left( {\frac{2\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}[/tex]

But you cannot say now that:
[tex]\sin \left( {\frac{\pi }{4} + \frac{{2\pi }}{3}} \right) = \sin \left( {\frac{\pi }{4}} \right) + \sin \left( {\frac{{2\pi }}{3}} \right) = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2}[/tex]

That's wrong, you have to use [tex]\sin \left( {\alpha + \beta } \right) = \sin \left( \alpha \right) \cdot \cos \left( \beta \right) + \cos \left( \alpha \right) \cdot \sin \left( \beta \right)[/tex]

Just fill in the formula :smile:
 
  • #11
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so by plugging in cos....

I come up with:
cos11pi/12=-sq rt2/4(sq rt3+1)
 
  • #12
TD
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That seems correct, [tex]\cos \left( {\frac{{11\pi }}{{12}}} \right) = - \frac{{\sqrt 2 }}{4}\left( {\sqrt 3 + 1} \right)[/tex]
 
  • #13
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Again, thank you for your assistance and patience.

:smile:
 
  • #14
TD
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No problem, you were on the right track for a longer time but I didn't understand your notation at first hehe.
 

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