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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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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

  2. anonymous
    • one year ago
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    Ok, but how can I get the answers like the ones I was given?

  3. anonymous
    • one year ago
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    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.

  4. anonymous
    • one year ago
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    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

  5. anonymous
    • one year ago
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    actually hold up

  6. anonymous
    • one year ago
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    ok so you used the \(\sin(a + b) = \sin a \cos b + \cos a \sin b \) identity?

  7. anonymous
    • one year ago
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    Yes, I think so.

  8. anonymous
    • one year ago
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    what did you get when you did it?

  9. anonymous
    • one year ago
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    Well, I only have right now, 2pi/3 + pi/4

  10. anonymous
    • one year ago
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    you have to substitute the values you got into the equation and evaluate it\[\sin \frac{ 11\pi }{ 12 }=\sin(\frac{ 2\pi }{ 3 }+\frac{ \pi }{ 4 })=\sin \frac{ 2\pi }{ 3 }\cos \frac{ \pi }{ 4}+\cos \frac{ 2\pi }{ 3 }\sin \frac{ \pi }{ 4 }\]

  11. anonymous
    • one year ago
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    Ok, but how can I get it to the root of a root form?

  12. anonymous
    • one year ago
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    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

  13. anonymous
    • one year ago
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    So, I'm still a little confused what to do with sin2π3cosπ4+cos2π3sinπ4

  14. anonymous
    • one year ago
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    what's sin (2pi/3)? Use your unit circle?

  15. anonymous
    • one year ago
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    Sq. Rt3/2?

  16. anonymous
    • one year ago
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    right. Now do cos π/4

  17. anonymous
    • one year ago
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    Sqrt2/2

  18. anonymous
    • one year ago
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    so far we have (√3)/2 * (√2)/2

  19. anonymous
    • one year ago
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    Oh I get it.

  20. anonymous
    • one year ago
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    great :). now do the other 2

  21. anonymous
    • one year ago
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    Yep thank you!!!

  22. anonymous
    • one year ago
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    you're welcome

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