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chels

  • 5 years ago

sin(2x)cos(5x) + cos(2x)sin(5x) use identities to simplify

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  1. myininaya
    • 5 years ago
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    sin(7x)

  2. chels
    • 5 years ago
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    how did you get that?

  3. myininaya
    • 5 years ago
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    sin(2x+5x)

  4. myininaya
    • 5 years ago
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    remember sin(z+y)=sinzcosy+sinycosz where in this case z=2x and y=5x

  5. chels
    • 5 years ago
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    where did the cos go?

  6. myininaya
    • 5 years ago
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    its an identity

  7. chels
    • 5 years ago
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    this is so confusing.. :(

  8. myininaya
    • 5 years ago
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    do you know the identity sin(z+y)=sinzcosy+sinycosz

  9. chels
    • 5 years ago
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    yes

  10. chels
    • 5 years ago
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    but i don't understand how it's used in this question because sin and cos are being multiplied together.

  11. myininaya
    • 5 years ago
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    ok what if you let z=2x and y=5x what would you have?

  12. chels
    • 5 years ago
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    i understand how to use the identity, i'm just confused as to where it's being used

  13. myininaya
    • 5 years ago
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    ok we have [sin(2x)][cos(5x)]+[cos(2x)][sin(2x)]=[sin(z)][cos(y)]+[cos(y)][sin(z)]=sin(y+z) where y=2x and z=5x so this =sin(2x+5x)=sin(7x)

  14. myininaya
    • 5 years ago
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    oops i wrote the samething twice and that post

  15. chels
    • 5 years ago
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    ohhh my gosh. i get it now

  16. chels
    • 5 years ago
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    i'm so sorry. haha

  17. myininaya
    • 5 years ago
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    ok good i don't have to write it over correctly then?

  18. chels
    • 5 years ago
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    no, that is good 1 / 1-cos(x) + 1 / 1 + cos(x)

  19. myininaya
    • 5 years ago
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    i assume that (1-cosx) and (1+cosx) are in the denominator right?

  20. chels
    • 5 years ago
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    yes

  21. myininaya
    • 5 years ago
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    combine those fractions

  22. chels
    • 5 years ago
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    okay

  23. myininaya
    • 5 years ago
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    what do you get now?

  24. myininaya
    • 5 years ago
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    numerator=(1+cosx)+(1-cosx)=2 and denominator=1-(cosx)^2=(sinx)^2

  25. myininaya
    • 5 years ago
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    2/(sinx)^2=2(cscx)^2

  26. chels
    • 5 years ago
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    i got a denominator of 1-cos(x)1+cos(x)

  27. myininaya
    • 5 years ago
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    (1-cosx)(1+cosx) use foil here

  28. chels
    • 5 years ago
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    ohh okay

  29. myininaya
    • 5 years ago
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    or you could use the fact that (a-b)(a+b)=a^2-b^2

  30. chels
    • 5 years ago
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    we haven't learned that one, but i'll note it

  31. myininaya
    • 5 years ago
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    thats an algebra thing it should had been algebra you learned that not trig

  32. chels
    • 5 years ago
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    hmm, i don't remember it thankyou though.

  33. myininaya
    • 5 years ago
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    (x-3)(x+3)=x^2-9 and (x-5)(x+5)=x^2-25 and so on....

  34. chels
    • 5 years ago
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    \[\tan(\pi/5) + \tan(\pi/10) = numerator. 1 - \tan(\pi/5)\tan(\pi/10)\]

  35. chels
    • 5 years ago
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    sorry. let me re-write that

  36. myininaya
    • 5 years ago
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    what you got the one before this one?

  37. chels
    • 5 years ago
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    numerator = tan(pi/5) + tan(pi/10) denominator = 1 - tan(pi/5)tan(pi/10)

  38. chels
    • 5 years ago
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    yes, i understood it.

  39. myininaya
    • 5 years ago
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    Nice!

  40. myininaya
    • 5 years ago
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    Do you remember the idenity tan(x+y)=[tanx+tany]/[1-tanxtany]

  41. chels
    • 5 years ago
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    yes

  42. myininaya
    • 5 years ago
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    can use that here?

  43. myininaya
    • 5 years ago
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    x=pi/5 and y=pi/10

  44. chels
    • 5 years ago
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    3pi / 10?

  45. chels
    • 5 years ago
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    tan 3pi/10, i mean

  46. myininaya
    • 5 years ago
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    yes!!! :))

  47. myininaya
    • 5 years ago
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    ok if you want more help, i'm gonna be away from my computer so people will notice you you might want to start a new thread k? if you like me become a fan

  48. chels
    • 5 years ago
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    okay

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