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Jamierox4ev3r

  • one year ago

If sin(x) = 1/3 and sec(y) = 5/4, where x and y lie between 0 and π/2, evaluate sin(x + y).

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  1. Jamierox4ev3r
    • one year ago
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    @ganeshie8 I actually have no idea what's going on here, so if you're online, I would love your help. thanks

  2. ganeshie8
    • one year ago
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    recall the angle sum identity for sin : \[\sin(x+y)=\sin x\cos y + \cos x\sin y\]

  3. ganeshie8
    • one year ago
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    basically you need to find those four pieces : \(\sin x, \cos x, ~~\sin y, \cos y\) then plug them in

  4. Jamierox4ev3r
    • one year ago
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    Alright, so I'm assuming that it would look a little like this: \(\sin (x + y) = (\frac{1}{3} )(\frac{5}{4})+ \sin(y)\cos(x) \)

  5. Jamierox4ev3r
    • one year ago
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    that's all we appear to know at this moment in time

  6. ganeshie8
    • one year ago
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    you have the correct idea but we don't really know the value of \(\cos y\) yet, what we're given is \(\sec y\) and not \(\cos y\)

  7. ganeshie8
    • one year ago
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    so we need to find \(\cos y\), \(\sin y\) and \(\cos x\) somehow from the given info

  8. Jamierox4ev3r
    • one year ago
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    oh whoops, completely forgot about the secant ;-;

  9. Jamierox4ev3r
    • one year ago
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    riiight

  10. ganeshie8
    • one year ago
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    its easy, lets find them

  11. Jamierox4ev3r
    • one year ago
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    Fair enough. Let's do it

  12. ganeshie8
    • one year ago
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    do you like triangles we're gonna need bunch of them

  13. Jamierox4ev3r
    • one year ago
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    triangles are good with me

  14. ganeshie8
    • one year ago
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    |dw:1441088950452:dw|

  15. ganeshie8
    • one year ago
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    try finding that missing side

  16. Jamierox4ev3r
    • one year ago
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    that would be \(\sqrt{8}\), I believe

  17. ganeshie8
    • one year ago
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    Yes! look at that triangle again, \(\cos x = ?\)

  18. Jamierox4ev3r
    • one year ago
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    \(\large\frac{\sqrt{8}}{3}\)

  19. ganeshie8
    • one year ago
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    Awesome! just draw another triangle using \(\sec y = 5/4\) and find whatever you need

  20. ganeshie8
    • one year ago
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    maybe let me draw the triangle for you

  21. Jamierox4ev3r
    • one year ago
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    I think I got this

  22. ganeshie8
    • one year ago
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    good, il just wait then :)

  23. Jamierox4ev3r
    • one year ago
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    let me try drawing it, just verify if I'm right. And thank you c:

  24. Jamierox4ev3r
    • one year ago
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    |dw:1441089382675:dw|

  25. Jamierox4ev3r
    • one year ago
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    oh and i forgot to draw in the right angle. but that's kind of a given

  26. Jamierox4ev3r
    • one year ago
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    this actually works out really well, the missing side would simply be 3

  27. ganeshie8
    • one year ago
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    yes, also the angle is \(y\), not \(x\) |dw:1441089486688:dw|

  28. Jamierox4ev3r
    • one year ago
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    oh right, whoops . Since this time around we're solving for the y

  29. Jamierox4ev3r
    • one year ago
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    I understand

  30. ganeshie8
    • one year ago
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    good, find the stuff you need and see if you can finish it off!

  31. Jamierox4ev3r
    • one year ago
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    |dw:1441089589409:dw|

  32. Jamierox4ev3r
    • one year ago
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    sounds good to me :)

  33. ganeshie8
    • one year ago
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    do let me know the final answer incase if you want me double check

  34. Jamierox4ev3r
    • one year ago
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    for sure, I want to make sure I'm doing the right thing. So \(\sin y = \frac{3}{5}\) and \(\sin x = \frac{4}{5}\) then we plug all of that back in to the identity \(\large sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5})\)

  35. Jamierox4ev3r
    • one year ago
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    Can this be simplified by multiplying the fractions together?

  36. Jamierox4ev3r
    • one year ago
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    or would this be the final form?

  37. ganeshie8
    • one year ago
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    for sure, I want to make sure I'm doing the right thing. So \(\sin y = \frac{3}{5}\) and \(\color{red}{\cos y }= \frac{4}{5}\) then we plug all of that back in to the identity \(\large sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5})\)

  38. Jamierox4ev3r
    • one year ago
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    right xD I'm sort of spacing out, but that's what I have in my notebook. Totally did not type that right

  39. Jamierox4ev3r
    • one year ago
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    if you looked at how I plugged things into the identity, I think it illustrates that I have a good understanding now of what's going on

  40. Jamierox4ev3r
    • one year ago
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    oh wait, and isn't it possible for \(\sqrt{8}\) to be be simplified to \(2\sqrt{2}\)?

  41. ganeshie8
    • one year ago
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    Ahh I see it was just a typo \(\large \sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5}) = \dfrac{2(2+3\sqrt{2})}{15}\) i would leave it at that

  42. Jamierox4ev3r
    • one year ago
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    Yeah, that seems to be the nicest form. I got it! Thank you so much. My goodness, the calculus professor said that this was all supposed to be review. Granted, much of it is, but some of these things I feel like I've never learned before. You're a life saver, thanks a bunch!

  43. ganeshie8
    • one year ago
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    np btw you're doing great! I can tell, you will have no problem reviewing/learning these basic trig/functions stuff as you continue with ur calc course... just dont let these stop you from learning calc... its okay to keep coming back to these topics when you find something not so clear in ur calc course..

  44. Jamierox4ev3r
    • one year ago
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    Yep, that's why I'm making sure I get all the review out of the way. I'm already gaining a much better understanding, so thank you very much for your efforts. I know that all my dedication to making sure I have the basics down will pay off in the long run. An thank you also for your kind words :') It's getting pretty late for me, but I have one more question I need to ask before I leave for the day. You wouldn't mind helping me with that one as well, would you?

  45. ganeshie8
    • one year ago
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    Ahh sure, ask..

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