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anonymous

  • one year ago

Will anybody explain to me how Sine Cosine and Tangent works? I missed school when they taught us, and when they tried to explain it to me I became brutally lost. :)

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  1. zepdrix
    • one year ago
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    Our trigonometric functions allow us to relate `the angle` of a triangle to its `sides`.

  2. zepdrix
    • one year ago
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    We have this clever acronym for remembering the relationships:\[\Large\rm \color{red}{\text{Soh}}\color{green}{\text{Cah}}\color{royalblue}{\text{Toa}}\] The `sine` of an angle is equivalent to the ratio of the `opposite` side to the `hypotenuse`. That's what the `o` and `h` stand for.\[\large\rm \color{red}{\sin x=\frac{opposite}{hypotenuse}}\]

  3. zepdrix
    • one year ago
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    Let's first look at a right triangle, and make sure we understand how to label the sides.

  4. zepdrix
    • one year ago
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    |dw:1444603738782:dw|So if this is my triangle, with angle x labeled here. Do you know which side to label as your `hypotenuse`?

  5. anonymous
    • one year ago
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    The side that is on top of the x, aka the "long side"

  6. zepdrix
    • one year ago
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    Good. The longest side. Another way I like to think of it, is in relation to the `right angle`. It's always the side `opposite the right angle`.|dw:1444604392651:dw|

  7. zepdrix
    • one year ago
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    Hmm with that idea in mind... Which side do you think we would label as being located `opposite angle x`?

  8. anonymous
    • one year ago
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    The side that is vertical. all the way to the right.

  9. zepdrix
    • one year ago
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    |dw:1444604529049:dw|

  10. zepdrix
    • one year ago
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    |dw:1444604552006:dw|Ok great. So the last side we label as being `adjacent` or `next to` or angle x.

  11. zepdrix
    • one year ago
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    |dw:1444604621344:dw|So if I have a particular triangle like this one, based on our definition of sin x, do you have an idea of how to find it using these values? :)

  12. anonymous
    • one year ago
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    Inverse Cosine?

  13. zepdrix
    • one year ago
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    No no, you're getting too fancy. Using something like Inverse Sine would allow us to figure out our angle x, yes. But we weren't up to that point yet XD We just want \(\large\rm \sin x=?\)

  14. anonymous
    • one year ago
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    I thought you said to find the angle :P so we want sin(x) which sine is Soa so Opposite over Adjacent?

  15. zepdrix
    • one year ago
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    Woooops :O\[\Large\rm \color{red}{\text{Soh}}\color{green}{\text{Cah}}\color{royalblue}{\text{Toa}}\]Soh, not Soa ya silly billy >.<

  16. anonymous
    • one year ago
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    Oh my lord x_x How did i even pass math class... SO! It'd be Opposite over Hypotenuse, yes?

  17. zepdrix
    • one year ago
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    Ok good.\[\large\rm \color{red}{\sin x=\frac{opposite}{hypotenuse}}\]Based on the way we labeled these sides, it looks like we're using the 4 and 5, ya?\[\large\rm \color{red}{\sin x=\frac{4}{5}}\]

  18. anonymous
    • one year ago
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    Yes yes. That's what the triangle tells us.

  19. zepdrix
    • one year ago
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    And ya, if we wanted to solve for the angle, we could apply the inverse sine function,\[\large\rm \arcsin\left(\sin x\right)=\arcsin\left(\frac{4}{5}\right)\]On the left you the composition of a function and it's inverse, which gives us back the argument,\[\large\rm x=\arcsin\left(\frac{4}{5}\right)\]

  20. zepdrix
    • one year ago
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    |dw:1444605278799:dw|If that's too confusing, another way to think of it is... When you change from sine to inverse sine, you just switch the stuff,

  21. anonymous
    • one year ago
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    .O. ohh my... but one question... So i was told if you do normal sine, and you go to find the hypotenuse, your equasion would look something like this: 5/sin(x)= adjacent? Why is this?

  22. anonymous
    • one year ago
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    *find the adjacent

  23. zepdrix
    • one year ago
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    |dw:1444605485213:dw|Ok let's see if we can do something with this triangle.

  24. anonymous
    • one year ago
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    *dies* x_x okayyy umm... SOH... 4/sin(30degrees) which will give us the adjacent...

  25. anonymous
    • one year ago
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    Opposite. it gives us the opposite.

  26. zepdrix
    • one year ago
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    Yah let's ignore the adjacent for now :) If we want to solve for \(\large y\),the opposite side, Then we want to pay attention to THIS stuff,|dw:1444605655180:dw|

  27. anonymous
    • one year ago
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    Yes. Sin(30)= y/4

  28. zepdrix
    • one year ago
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    Mmm k good. And how would you `isolate` the y? You want to solve for y, so you need to get it alone somehow. It's being `divided` by 4 right now. How can we undo that?

  29. anonymous
    • one year ago
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    multiply it by 4. which means we have to do it to the other side so 4times sin(30) = y

  30. zepdrix
    • one year ago
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    Good. And keep in mind that this 30 is like... locked in the sine function, he can't interact with the 4 in any way.\[\large\rm 4\sin(30)\ne \sin(4\cdot30)\]

  31. zepdrix
    • one year ago
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    \[\large\rm y=4\sin(30)\]And you would just use your calculator to finish that one off, unless of course you've learned about your 30/60/90 triangle, in which case you might be able to do it without a calculator.

  32. anonymous
    • one year ago
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    So then we'd do sin(30) first. then multiply the answer by 4?

  33. anonymous
    • one year ago
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    I got 2 when i plugged it into my calculator.

  34. zepdrix
    • one year ago
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    Yay good job! How would solve for the adjacent side, x? Let's pretend that we don't know what y is. So we can make sure of our cosine definition.

  35. zepdrix
    • one year ago
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    |dw:1444606171089:dw|Again, we don't want more variables than x. So we're dealing with these three pieces of information, ya?

  36. anonymous
    • one year ago
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    Alright... so CAH... Cos(30)= x/4

  37. anonymous
    • one year ago
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    and with that we'd need x alone so we multiply it by both sides correct?

  38. anonymous
    • one year ago
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    4 is it XD oops.

  39. zepdrix
    • one year ago
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    \[\large\rm \cos(30)=\frac{x}{4}\]Multiply 4? Ok seems like a good idea:\[\large\rm 4\cos(30)=\frac{x}{\cancel4}\cdot\cancel4\]\[\large\rm 4\cos(30)=x\]And unfortunately, this one is going to work out to a weird decimal length, but that's ok.

  40. anonymous
    • one year ago
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    I got 3.46 for x o.o

  41. zepdrix
    • one year ago
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    Here is a nice way to check your work: Recall that your hypotenuse should be the `longest side`. 3.46 is shorter than 4 \(\large\rm \color{green}{\checkmark}\) I tried to also draw that angle somewhat accurately to be a 30 degree angle. So if that vertical length is 2, would 3.5 be about right for the bottom length? yaaa that's prolly right!

  42. anonymous
    • one year ago
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    Seems easy enough to remember.

  43. zepdrix
    • one year ago
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    |dw:1444606546042:dw|Let's try this problem. We need to find the length of the hypotenuse, z.

  44. anonymous
    • one year ago
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    Cos(30)= 5/z

  45. zepdrix
    • one year ago
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    \[\large\rm \cos(30)=\frac{5}{z}\]Ok good. Hmmm, notice our z is in the denominator, this makes things a little trickier. Any ideas? :o

  46. anonymous
    • one year ago
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    Uhhh.... divide five by both sides?...

  47. anonymous
    • one year ago
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    5/cos(30)= z

  48. zepdrix
    • one year ago
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    Ya let's try that:\[\large\rm \frac{1}{5}\cdot\cos(30)=\frac{\cancel5}{z}\cdot\frac{1}{\cancel5}\]The z is still stuck in the denominator though! Ok you had the good sense to flip it after that though?\[\large\rm \frac{\cos(30)}{5}=\frac{1}{z}\qquad\to\qquad \frac{5}{\cos(30)}=z\]

  49. zepdrix
    • one year ago
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    Or maybe you got lucky :) I'm not sure which lol

  50. anonymous
    • one year ago
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    XD im pretty sure i got lucky... i didn't even realize i skipped a whole piece.

  51. zepdrix
    • one year ago
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    When your variable is stuck in the denominator, these are the steps I would recommend:\[\large\rm \cos(30)=\frac{5}{z}\]Multiply both sides by z,\[\large\rm z\cos(30)=5\]Divide both sides by cos(30),\[\large\rm z=\frac{5}{\cos(30)}\]

  52. zepdrix
    • one year ago
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    So when your variable is stuck in the bottom, the process is a little bit different. Just a bit trickier.

  53. anonymous
    • one year ago
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    Indeed. When I plugged the equation in, I got 3.8

  54. zepdrix
    • one year ago
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    Ok, again let's check our work. The hypotenuse should be the longest side. But our adjacent side is 5. Uh oh! 3.8 < 5

  55. anonymous
    • one year ago
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    My second time i got 5.8 when i did Cos(30), got my answer, then divided 5 by my answer (5/ ans is what im trying to say...)

  56. zepdrix
    • one year ago
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    Do you have a calculator that uses the "Ans" thing?

  57. anonymous
    • one year ago
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    Yes i do.

  58. zepdrix
    • one year ago
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    Ya I really like that feature. You can do the problem all at once, \(\large\rm 5\div\cos(30)\) Or in parts as you described, \(\large\rm \cos(30)\quad \boxed{=}\) \(\large\rm 5\div Ans \quad \boxed{=}\) Whichever way makes more sense to you :)

  59. anonymous
    • one year ago
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    They both do. But sometimes double checking prevents me from making a mistake, gladly.

  60. zepdrix
    • one year ago
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    I need a math break I think :) lol Too much maf. Here is one more you can work on though.

  61. zepdrix
    • one year ago
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    |dw:1444607625616:dw|

  62. zepdrix
    • one year ago
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    So if you get bored and wanna try another one, try to solve for x in that triangle :)

  63. zepdrix
    • one year ago
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    I'mma go make some foods a sec >.<

  64. anonymous
    • one year ago
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    Alright. I'll attempt to figure this one out. and you enjoy your food. I need a math break after this too XD I have alot more homework... lol

  65. anonymous
    • one year ago
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    Tan(41)=7/x 7/tan(41)=x 7/tan(41)= 8.1 That's what I got.

  66. anonymous
    • one year ago
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    Lol then i got 10.7 for the hypotenuse using the same opposite.

  67. zepdrix
    • one year ago
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    7/tan(41)=8.05, so ya 8.1 sounds good. 10.7? 0_o weird

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