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saifoo.khan

  • 2 years ago

Easy trig problem below:

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  1. jasonxx
    • 2 years ago
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    ?

  2. saifoo.khan
    • 2 years ago
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    \[\frac{3A}{sinA} + \frac{\cos3A}{\cos A}≡ 4 \cos 2A\]

  3. jasonxx
    • 2 years ago
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    cos(3A) = cos(2A+A) =4 cos³ A - 3 cos A

  4. jasonxx
    • 2 years ago
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    taking the LCM and then using 2 sinA * cos A= sin 2A

  5. jasonxx
    • 2 years ago
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    and cos 2A= cos^2 2A -1

  6. saifoo.khan
    • 2 years ago
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    I'm not sure about your working but i completed till here: \[\frac{\cos A \sin 3A + \cos 3A \sin A}{\sin A \cos A}\]

  7. saifoo.khan
    • 2 years ago
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    Then: \[\frac{\sin(A+3A)}{\sin A \cos A}\]idk what to do after this..

  8. saifoo.khan
    • 2 years ago
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    @jasonxx ?

  9. jasonxx
    • 2 years ago
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    well, i guess expand everything

  10. zepdrix
    • 2 years ago
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    \[\frac{\sin(4A)}{\frac{1}{2}\sin(2A)}\] This the next step? :) hmm fun problem.

  11. zepdrix
    • 2 years ago
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    \[\frac{4\sin(2A) \cos(2A)}{\sin(2A)}\] \[=4\cos(2A)\] Something like that... :D Yayyy team!

  12. saifoo.khan
    • 2 years ago
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    HOW? HOW? HOW?

  13. jasonxx
    • 2 years ago
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    sin(4A)=sin(2* 2A)

  14. zepdrix
    • 2 years ago
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    Oh like say what each step that's being performed? :D

  15. saifoo.khan
    • 2 years ago
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    Last 3 steps, yes sir. @zepdrix ;D

  16. jasonxx
    • 2 years ago
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    and sin (2*2A) is like sin (2* X)= 2*2 sin (2A)cos(2A)

  17. zepdrix
    • 2 years ago
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    Hmm so we're taking advantage of Sine's Double Angle Formula, On top, in one direction, on the bottom in the other direction. Yah looks like Jason has the right idea also ^^

  18. jasonxx
    • 2 years ago
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    :)) thanks @zepdrix

  19. saifoo.khan
    • 2 years ago
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    I have this general formula with me: sin2A = 2sinAcosA But how did you guys converted sin4A into this?

  20. jasonxx
    • 2 years ago
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    Let's say A=X then its gona be sin (2X), now X=2A therefore its gona be the conversion mentioned above

  21. saifoo.khan
    • 2 years ago
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    I think i got it now. Thanks @jasonxx and @zepdrix . :)

  22. jasonxx
    • 2 years ago
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    :) keep rockin

  23. zepdrix
    • 2 years ago
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    Oh sorry. I wasn't going into much detail because I thought you were testing us :) lol

  24. saifoo.khan
    • 2 years ago
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    Lol. No. this was the last problem of 26. i had the answers but wasn't getting the last step!

  25. saifoo.khan
    • 2 years ago
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    @zepdrix Where did the 1/2 in your second last step go? D:

  26. zepdrix
    • 2 years ago
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    So we're dividing by a fraction, let's fix that a sec. \[\frac{\sin(4A)}{\frac{1}{2}\sin(2A)} \quad = \quad \frac{2\sin(4A)}{\sin(2A)}\] Then we apply the Sine Double Angle Formula to the top part.\[\frac{2\sin(2\cdot 2A)}{\sin(2A)} \quad = \quad \frac{2\cdot 2 \sin(2A)\cos(2A)}{\sin(2A)}\]

  27. saifoo.khan
    • 2 years ago
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    Oh! Now it makes perfect sense to me!

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