- #1

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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?

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- Thread starter TonyC
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- #1

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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?

- #2

TD

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How did you get that? Does't seem right to me.

- #3

Tide

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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=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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This is correct: [tex]\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}[/tex]TonyC said:11pi/12=165 degrees

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

sq root 3/3+sq root 2/2

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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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

- #11

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so by plugging in cos....

I come up with:

cos11pi/12=-sq rt2/4(sq rt3+1)

I come up with:

cos11pi/12=-sq rt2/4(sq rt3+1)

- #12

TD

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- #13

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Again, thank you for your assistance and patience.

- #14

TD

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