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anonymous

  • one year ago

Check my work if I'm correct. Find the exact value of sin(-11pi/12).

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  1. anonymous
    • one year ago
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    So I changed it from radians to degrees to make it easier.

  2. hybrik
    • one year ago
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    sin(-11(3.1415926535)/12)

  3. anonymous
    • one year ago
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    I'm not allowed to use the calculator. :)

  4. hybrik
    • one year ago
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    = sin(0.26)

  5. hybrik
    • one year ago
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    math I just remember first digits of pi so

  6. hybrik
    • one year ago
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    You would just simplify 3.141592/12, which is around 0.26

  7. hybrik
    • one year ago
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    then find the sin of 0.26

  8. anonymous
    • one year ago
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    Should I use a reference angle to find -165?

  9. anonymous
    • one year ago
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    So it'll be 15.

  10. anonymous
    • one year ago
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    |dw:1438110805222:dw|

  11. anonymous
    • one year ago
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    \[= -(\sin60\cos45-\cos60\sin45)\]

  12. Nnesha
    • one year ago
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    i would change negative angle to positive and then use one of de formula

  13. Nnesha
    • one year ago
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    i like positive stuff :D

  14. anonymous
    • one year ago
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    |dw:1438110888230:dw|

  15. anonymous
    • one year ago
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    |dw:1438110954725:dw|

  16. anonymous
    • one year ago
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    I don't know if I did it right, so what do you guys think?

  17. Nnesha
    • one year ago
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    gimme a sec

  18. anonymous
    • one year ago
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    All right. :)

  19. Nnesha
    • one year ago
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    okay so i add 360 int -165 360 -165 =195 i used sin(a+b) formula \[\huge\rm sin(a+b) =\sin A \times \cos B + \cos A \times \sin B\] \[\sin(45+150) =\sin 45 \cos 150 + \cos 45 \times \sin 150\] \[\sin(45+150) =\frac{ \sqrt{2} }{ 2} \times \frac{ -\sqrt{3} }{ 2 }+\frac{ \sqrt{2} }{ 2 }\times \frac{ 1 }{ 2 }\] and got \[\frac{ -\sqrt{6} + \sqrt{2 }}{ 4}\] so your answer is correct :D

  20. anonymous
    • one year ago
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    Thank you! :)

  21. Nnesha
    • one year ago
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    gO_Od job! i almost forgot these stuff ;D

  22. freckles
    • one year ago
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    the cool thing about these is your start doesn't have to match someone elses start that like I would have done it this way: \[\sin(\frac{-11\pi}{12}) \\ \text{ sine function is odd } \\ -\sin(\frac{11\pi}{12}) \\ -\sin(165^o) \\ -\sin(120^o+45^o) \\ -[\sin(120^o)\cos(45^o)+\sin(45^o) \cos(120^o)] \\ -[\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \frac{-1}{2}] \\ -[\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}] \\ -[\frac{\sqrt{6}-\sqrt{2}}{4}] \\ \text{ distribute outside negative } \\ \frac{-\sqrt{6}--\sqrt{2}}{4} \\ \frac{-\sqrt{6}+\sqrt{2}}{4}\]

  23. Nnesha
    • one year ago
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    yeah but i wasn't sure that we can take out negative sign \[\sin(\frac{ -11\pi }{ 2 }) --> -\sin(\frac{ 11\pi }{ 2 })\]

  24. Nnesha
    • one year ago
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    ik sin(-x)=-sin(x) hmm so that's possible

  25. freckles
    • one year ago
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    right and if we had cos(-x) then this is just cos(x) since cos is even

  26. freckles
    • one year ago
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    that is for example \[\cos(\frac{-11\pi}{12})=\cos(\frac{11\pi}{12})\]

  27. Nnesha
    • one year ago
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    true :=)

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