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lilsis76

  • 2 years ago

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

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  1. lilsis76
    • 2 years ago
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    |dw:1350681532092:dw|

  2. phi
    • 2 years ago
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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|

  3. phi
    • 2 years ago
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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?

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

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

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

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

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

  9. phi
    • 2 years ago
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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)

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

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

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

  13. lilsis76
    • 2 years ago
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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?

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

  15. phi
    • 2 years ago
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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)?

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

  17. phi
    • 2 years ago
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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)

  18. lilsis76
    • 2 years ago
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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

  19. phi
    • 2 years ago
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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)?

  20. phi
    • 2 years ago
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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)

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

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

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

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

  25. lilsis76
    • 2 years ago
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    okay

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

  27. lilsis76
    • 2 years ago
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    thats all im understanding

  28. phi
    • 2 years ago
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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

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

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

  31. phi
    • 2 years ago
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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.

  32. lilsis76
    • 2 years ago
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    okay, thank you

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

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

  35. phi
    • 2 years ago
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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

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

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

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

  39. phi
    • 2 years ago
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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)

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

  41. lilsis76
    • 2 years ago
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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?

  42. phi
    • 2 years ago
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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.

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

  44. phi
    • 2 years ago
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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 \[ \sin(2x)= \frac{4\sqrt{6}}{25} \]

  45. lilsis76
    • 2 years ago
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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

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