Sine, cosine and tangency of angle 11pi/12

[TonyC]

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?

 

[TD]

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

 

[Tide]

How did you arrive at those?

For the first one i get

$$sin \frac {11 \pi}{12} = \frac {\sqrt 2}{4} (\sqrt 3 - 1)$$

 

[TonyC]

I started by working in radians.

 

[Tide]

And then what?

 

[TonyC]

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

 

[TD]

11pi/12=165 degrees

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

This is correct: $$\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}$$

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 $\sin \left( {\alpha + \beta } \right) = \sin \left( \alpha \right) \cdot \cos \left( \beta \right) + \cos \left( \alpha \right) \cdot \sin \left( \beta \right)$

 

[TonyC]

I am still confused...

 

[TonyC]

P.S. what program are you using so I don't have to keep writing out the equations long hand?

 

[TD]

Well, we have that $$\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}$$

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

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

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

Just fill in the formula

 

[TonyC]

so by plugging in cos...

I come up with:
cos11pi/12=-sq rt2/4(sq rt3+1)

 

[TD]

That seems correct, $$\cos \left( {\frac{{11\pi }}{{12}}} \right) = - \frac{{\sqrt 2 }}{4}\left( {\sqrt 3 + 1} \right)$$

 

[TonyC]

Again, thank you for your assistance and patience.

 

[TD]

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