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anonymous

  • one year ago

If sin theta=3/5 and theta is in quad 2 the exact form of sin (theta+ pi/6) is .....? literally have not figured out this question at all..

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  1. anonymous
    • one year ago
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    I dont understand what you are asking, what do you mean by 'quad 2'.

  2. anonymous
    • one year ago
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    14mdaz, you have different quadrants when dealing with the unit circle and that is what he/she is referring to

  3. anonymous
    • one year ago
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    oh quadrants

  4. anonymous
    • one year ago
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    quadrant 2 is between pi/2 and pi

  5. anonymous
    • one year ago
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    What I don't get is what he / she is wanting from the exact form of sin (theta+ pi/6) is

  6. anonymous
    • one year ago
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    it will not be the cleanest value

  7. anonymous
    • one year ago
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    Is he wanting to add sin theta=3/5 + pi/6???

  8. anonymous
    • one year ago
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    no he wants theta plus pi/6 all in rads im guessing

  9. anonymous
    • one year ago
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    applied to sin function

  10. anonymous
    • one year ago
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    This whole question is just confusing me on top of that right now lol its asking for it to be in exact form

  11. anonymous
    • one year ago
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    Are you allowed to use a calculator or do you need to find \( \sin \theta = 3/5 \) without a calculator?

  12. anonymous
    • one year ago
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    Without a calculator sadly...

  13. anonymous
    • one year ago
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    hmm...

  14. anonymous
    • one year ago
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    |dw:1436649831637:dw|

  15. anonymous
    • one year ago
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    :3

  16. anonymous
    • one year ago
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    I think 14mdaz is explaining so I am going to step back

  17. anonymous
    • one year ago
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    uhh no i just doodled, its a highly inaccurate diagram

  18. anonymous
    • one year ago
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    I don't even think he knows

  19. zepdrix
    • one year ago
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    Still need help on this one shorty? :)

  20. anonymous
    • one year ago
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    Yes pleaseee!

  21. zepdrix
    • one year ago
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    You need to apply your Angle Sum Formula:\[\large\rm \sin(\alpha+\beta)=\sin \alpha \cos \beta+\sin \beta \cos \alpha\]

  22. zepdrix
    • one year ago
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    Let's apply that before we do anything else

  23. zepdrix
    • one year ago
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    \[\large\rm \sin\left(\theta+\frac{\pi}{6}\right)=\sin \theta \cos \frac{\pi}{6}+\sin\frac{\pi}{6}\cos \theta\]Do you understand how I applied that formula? :)

  24. anonymous
    • one year ago
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    Yeah I think I have done it that way and got the answer [3*sqrt(3) - 4]/10 but I don't think its right :/

  25. zepdrix
    • one year ago
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    Oooo yes good job! That looks correct!! :)

  26. anonymous
    • one year ago
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    But how would the work look like?

  27. anonymous
    • one year ago
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    But when I go to check my answer something comes out differently.

  28. anonymous
    • one year ago
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    sin(theta)cos(pi/6) + cos(theta)sin(pi/6) = (3/5)(sqrt(3)/2) - (4/5)(1/2)

  29. anonymous
    • one year ago
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    The work is correct right?

  30. zepdrix
    • one year ago
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    \[\large\rm \sin\left(\theta+\frac{\pi}{6}\right)=\left(\sin \theta\right)\left(\cos \frac{\pi}{6}\right)+\left(\sin\frac{\pi}{6}\right)\left(\cos \theta\right)\]\[\large\rm \sin\left(\theta+\frac{\pi}{6}\right)=\left(\frac{3}{5}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{-4}{5}\right)\]

  31. zepdrix
    • one year ago
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    Plugging in the pieces :) ya looks right

  32. anonymous
    • one year ago
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    But is there a way to check it?

  33. zepdrix
    • one year ago
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    Hmmm.

  34. anonymous
    • one year ago
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    He is using the Sum and Difference formula.

  35. zepdrix
    • one year ago
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    \[\large\rm \sin(\theta)=\frac{3}{5}\qquad\to\qquad \sin^{-1}\frac{3}{5}=\theta\approx0.6435\] So then the sine of that angle theta... plus pi/6 should approximately give us (3sqrt3-4)/10, whatever decimal that works out to. Kind of a tough problem to check your work on :)

  36. anonymous
    • one year ago
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    You can easily check with a calculator. Just plugin the values

  37. anonymous
    • one year ago
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    Oh alright cuz my professor was all like you can check your work which is why I was asking

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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