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lilsis76 Group Title

suppose sin theta = 1/5 and 0<theat< pi/2 compute cos theta and compute sin 2theta... i have these formulas given but i dont know what to do with them. let me draw them

  • 2 years ago
  • 2 years ago

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  1. lilsis76 Group Title
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    |dw:1350681532092:dw|

    • 2 years ago
  2. phi Group Title
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    i have these formulas given... The expressions you posted are specific examples, not formulas. To answer this question suppose sin x = 1/5 and 0<x< pi/2 (Let me use x, it's easier to type than theta) First, recall the definition of sin: opposite over hypotenuse. 2nd, it is helpful to draw a graph. |dw:1350682184808:dw|

    • 2 years ago
  3. phi Group Title
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    they tell us x is between 0 and pi/2 (90º) so we know x is in the first quadrant. the cos x is adjacent over hypotenuse. we know the hypotenuse (it's 5) use pythagorean theorem to find the bottom leg of the right triangle (that will be the adjacent side we need for the cos) can you do that?

    • 2 years ago
  4. lilsis76 Group Title
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    yes let me try it on a paper.

    • 2 years ago
  5. lilsis76 Group Title
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    a= sqrt 24 -> i can reduce giving me a=sqrt 4 sqrt 6 a= 2sqrt6

    • 2 years ago
  6. phi Group Title
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    yes, use a's value to find \[ \cos(x)= \frac{2\sqrt{6}}{5}\]

    • 2 years ago
  7. phi Group Title
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    |dw:1350682709676:dw|

    • 2 years ago
  8. lilsis76 Group Title
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    sorry about that, this thing is being slow, this openstudy site

    • 2 years ago
  9. phi Group Title
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    to find compute sin 2x I remember sin(a+b)= sin(a)cos(b)+cos(a)sin(b) if a and b are the same=x, we can write this as sin(x+x) = sin(x) cos(x) + cos(x) sin(x) sin(2x)= 2 sin(x) cos(x) so replace sin(x) with 1/5 and cos(x) with 2sqrt(6)/5 in that formula and simplify to get sin(2x)

    • 2 years ago
  10. phi Group Title
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    being slow Yes, sometimes it acts up. Some days are better than others.

    • 2 years ago
  11. lilsis76 Group Title
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    okay ill try to find the first one, but how did you get \[2\sqrt{6} / 5 ? <--- how did u get the 5?\]

    • 2 years ago
  12. phi Group Title
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    the cos x is adjacent over hypotenuse

    • 2 years ago
  13. lilsis76 Group Title
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    how did u get the 5? haha sorry i cant read the other writting either. and i did sin 2x= sin 2(1/5) - that would make it sin 2/5 right?

    • 2 years ago
  14. lilsis76 Group Title
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    okay i see the cos =5 now. is 2/5 right?

    • 2 years ago
  15. phi Group Title
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    no. it is not the simple (otherwise it would not be math) first, what do you have for sin(x) and cos(x)?

    • 2 years ago
  16. lilsis76 Group Title
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    sin x is opp/hyp= 1/5 cos x is adj/hyp= 2sqrt{6}/5

    • 2 years ago
  17. phi Group Title
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    ok, that is good. I repeat my post from above to find compute sin 2x I remember sin(a+b)= sin(a)cos(b)+cos(a)sin(b) if a and b are the same=x, we can write this as sin(x+x) = sin(x) cos(x) + cos(x) sin(x) sin(2x)= 2 sin(x) cos(x) so replace sin(x) with 1/5 and cos(x) with 2sqrt(6)/5 in that formula and simplify to get sin(2x)

    • 2 years ago
  18. lilsis76 Group Title
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    in the sin(a+b)= sin(a)cos(b)+cos(a)sin(b)? like sin( 1/5 + 2sqrt6/5) = sin 1/5 cos 2sqrt6/5 + cos 1/5 sin 2sqrt6/5

    • 2 years ago
  19. phi Group Title
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    no. don't get confused. I was showing that the "sum of angles" formula simplifies to sin(2x)= 2 sin(x) cos(x) For example, if x were 30º, you could use this formula to find the sin(60º) sin(60)= sin(2*30)= 2 sin(30) cos(30) in this case, you are asked to find the sin(2x). You know sin(x) and cos(x). Can you find sin(2x)?

    • 2 years ago
  20. phi Group Title
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    sin x is opp/hyp= 1/5 cos x is adj/hyp= 2sqrt{6}/5 replace sin(x) with 1/5 replace cos(x) with 2sqrt{6}/5 in the formula sin(2x)= 2 sin(x) cos(x)

    • 2 years ago
  21. lilsis76 Group Title
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    okay let me try it. i am trying to find it on my paper

    • 2 years ago
  22. lilsis76 Group Title
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    okay i got this then

    • 2 years ago
  23. lilsis76 Group Title
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    sin 2x = 2 sin 1/5 cos 2sqrt 6 / 5

    • 2 years ago
  24. phi Group Title
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    be careful x is not 1/5 sin(x) is 1/5 try again.

    • 2 years ago
  25. lilsis76 Group Title
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    okay

    • 2 years ago
  26. lilsis76 Group Title
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    sin 2x = 2sin 1/5 cos x

    • 2 years ago
  27. lilsis76 Group Title
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    thats all im understanding

    • 2 years ago
  28. phi Group Title
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    when we say sin x =1/5 and then write, for example cos(x) sin(x) we can replace sin(x) with 1/5 (that is what = means) so cos(x) sin(x) could be written as cos(x) * 1/5 can you use that idea in sin(2x)= 2 sin(x) cos(x) when sin x= 1/5 cos x= 2sqrt{6}/5

    • 2 years ago
  29. phi Group Title
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    in other words, replace sin x with 1/5 and replace cos(x) with 2sqrt(6)/5

    • 2 years ago
  30. lilsis76 Group Title
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    into the formula you gave me? the...sin(2x)= 2 sin(x) cos(x) ? what is that equation for also?

    • 2 years ago
  31. phi Group Title
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    yes, in the formula sin(2x)= 2 sin(x) cos(x) I'll post a video that explains this, so you can watch it when you have time.

    • 2 years ago
  32. lilsis76 Group Title
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    okay, thank you

    • 2 years ago
  33. lilsis76 Group Title
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    like this: 2(1/5) = 2(1/5)(2sqrt6 / 5) ?

    • 2 years ago
  34. lilsis76 Group Title
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    HEYY?! i just realized, did we solve the compute cos theta? was that the 2sqrt6 / 5?

    • 2 years ago
  35. phi Group Title
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    closer. we know sin(x) and cos(x). but we do not know sin(2x) we start with sin(2x)= 2 sin(x) cos(x) replace sin(x) with 1/5 sin(2x)= 2* 1/5 * cos(x) replace cos(x) with 2sqrt6/5 sin(2x)= 2* 1/5 * 2sqrt6/5 notice that sin(2x) is not 2*1/5

    • 2 years ago
  36. lilsis76 Group Title
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    hmmm....okay

    • 2 years ago
  37. lilsis76 Group Title
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    will the 2 (1/5) = 2/5? or is that wrong?

    • 2 years ago
  38. lilsis76 Group Title
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    i ddont know how to mult. fractions.

    • 2 years ago
  39. phi Group Title
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    yes 2*1/5 is 2/5 the answer is \[ \sin(2\theta)= \frac{4\sqrt{6}}{25} \] (I used theta again just to match the problem)

    • 2 years ago
  40. phi Group Title
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    Here is the first video on trig http://www.khanacademy.org/math/trigonometry/v/basic-trigonometry

    • 2 years ago
  41. lilsis76 Group Title
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    oh....okay, thank you ill watch the video. but one quesiton. why dont we multiply the sqrt by the 2 also?

    • 2 years ago
  42. phi Group Title
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    Here is a list of videos on fractions http://www.khanacademy.org/math/arithmetic/fractions I would watch the first few just to see if they make sense. Later on, if you run into a problem, go to this site and search for a video that explains it.

    • 2 years ago
  43. lilsis76 Group Title
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    okay. thank you so much. imma watch a few of these then get back to the homework.

    • 2 years ago
  44. phi Group Title
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    \[ \sin(2x)= 2\cdot \frac{1}{5} \cdot \frac{ 2\sqrt{6}}{5} \] to multiply fractions you multiply top*top and bottom*bottom (whole numbers are "tops") \[ \sin(2x)= \frac{2\cdot 1 \cdot 2\sqrt{6}}{5\cdot 5} \] that simplifies to \[ \sin(2x)= \frac{4\sqrt{6}}{25} \]

    • 2 years ago
  45. lilsis76 Group Title
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    oh...okay now i see it, dang lol i never saw it like that before. well thank you I shall watch the videos and ill be back later

    • 2 years ago
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