Could someone please explain how to find an exact value of sin(11pi/12)? I'm not sure how to get them in the answer as the ones given: A. (Sq. Rt 2 - Sq. Rt 6)/4 B. (/Sq. Rt 6 - Sq. Rt 2)/4 C. (Sq. Rt 6 - Sq. Rt 2)/4 D. (Sq. Rt 6 + Sq. Rt 2)/4

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Could someone please explain how to find an exact value of sin(11pi/12)? I'm not sure how to get them in the answer as the ones given: A. (Sq. Rt 2 - Sq. Rt 6)/4 B. (/Sq. Rt 6 - Sq. Rt 2)/4 C. (Sq. Rt 6 - Sq. Rt 2)/4 D. (Sq. Rt 6 + Sq. Rt 2)/4

Mathematics
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11pi/12 is half of 11pi/6, so you can use a half angle identity. \[\sin \frac{ \theta }{ 2 }=\sqrt{\frac{ 1-\cos \theta }{ 2 }}\] where Θ=11π/6
Ok, but how can I get the answers like the ones I was given?
Because I simplified the problem to: 2pi/3 + pi/4. But I'm not sure how to get them simplified to what answers I am given.

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Other answers:

you used a different identity, so you got the answer in a different format. The easiest way to get the answer in the root of a root format is to use the half angle identity
actually hold up
ok so you used the \(\sin(a + b) = \sin a \cos b + \cos a \sin b \) identity?
Yes, I think so.
what did you get when you did it?
Well, I only have right now, 2pi/3 + pi/4
you have to substitute the values you got into the equation and evaluate it\[\sin \frac{ 11\pi }{ 12 }=\sin(\frac{ 2\pi }{ 3 }+\frac{ \pi }{ 4 })=\sin \frac{ 2\pi }{ 3 }\cos \frac{ \pi }{ 4}+\cos \frac{ 2\pi }{ 3 }\sin \frac{ \pi }{ 4 }\]
Ok, but how can I get it to the root of a root form?
it looks like you don't have to do that, and if you use the unit circle to find those value it will give you the answer
So, I'm still a little confused what to do with sin2π3cosπ4+cos2π3sinπ4
what's sin (2pi/3)? Use your unit circle?
Sq. Rt3/2?
right. Now do cos π/4
Sqrt2/2
so far we have (√3)/2 * (√2)/2
Oh I get it.
great :). now do the other 2
Yep thank you!!!
you're welcome

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