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Solving trigonometric equations

  1. Jul 9, 2006 #1
    as we know that (sin(x))^2 + (cos(x))^2=1
    how about (sin(3x))^2 + (cos(3x))^2 and 5sin(3x)^2 + 6cos(3x)^2??

    how can we solve these problems??
    thanx
     
  2. jcsd
  3. Jul 9, 2006 #2
    What are you trying to solve? You could substitute numbers for the x's and compute the value.

    I don't believe there's a nifty identity for (sin(3x))^2 + (cos(3x))^2 even though it does look somewhat similar to (sin(x))^2 + (cos(x))^2.
     
  4. Jul 9, 2006 #3
    Actually, aslong as the arguments match it equals one. So (sin(3x))^2 + (cos(3x))^2 does equal one.
     
  5. Jul 9, 2006 #4

    Hootenanny

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    Indeed, if we apply the identity [itex]\sin^{2}\theta +\cos^{2}\theta = 1[/itex],to the above expression and simply use the substitution [itex]3x = \theta[/itex].... :wink:
     
  6. Jul 9, 2006 #5

    HallsofIvy

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    For 5sin(3x)^2 + 6cos(3x)^2, write it as 5sin^3(3x)+ 5cos^2(3x)+ cos^2(3x)= 5(sin^3(3x)+ cos^2(3x))+ cos^2(3x)= 5+ cos^2(3x).
     
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