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Trig Identities Applications Question

  1. Apr 18, 2012 #1
    Hello! I've been tackling the question 'Express sin3x+sinx as a product and hence solve 1/2(sin3x+sinx)=sin2x ; x∈R' but I'm stumped - I'm not sure whether I've even approached it correctly. This is what I did:

    sin(3x+x)=sin3x.cosx+sinx.cos3x

    inserting this into the second equation,
    1/2(sin3x.cosx+sinx.cos3x)=2sinx.cosx

    Moving cosx over
    4sinx=sin3x+3sinx

    sinx=sin3x

    sin(2x+x)-sinx=0

    sin2xcosx+cos2x.sinx-sinx=0

    2sinxcos^2x+(cos^2 x - sin^2 x) sinx - sinx = 0

    2sinxcos^2 x + cos^2xsinx - sin^3 x -sinx=0

    sinx(2cos^2 x + cos^2 x - sin^2 x - 1) = 0

    sinx(3cos^2x - sin^2x -1) =0

    BRACKETS:

    3cos^2x - sin^2x - 1 = 0

    3cos^2x - (1- cos^2x) - 1 = 0

    3cos^2x - 1 + cos^2x - 1 = 0

    4cos^2x - 2 = 0

    cos^2x = 1/2
    and
    sinx = 0

    But at this point I'm stumped. The answer says x = n∏/2 , and I would've gone about the second equation by doing:
    cosx = √(1/2)
    x = α = cos^-1(√(1/2)) but this is obviously wrong... :S Argh, how am I supposed to get that answer?

    Thanks in advance.
     
    Last edited: Apr 18, 2012
  2. jcsd
  3. Apr 18, 2012 #2
    1. Is the equation 1/2(sin3x + sinx) =sin2x ie do you have the 2 and 1 the wrong way round in the first line of your question?

    2. sin(3x) + sin(x) is not the same as sin(3x+x)

    This page might help you.
    http://www.sosmath.com/trig/prodform/prodform.html
     
  4. Apr 18, 2012 #3
    Ah, yes it is, sorry. And thanks, I'll see what else I can do.
     
  5. Apr 18, 2012 #4

    Curious3141

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  6. Apr 18, 2012 #5
    My working now:
    sin3x+sinx = 2sin(3x+x/2)cos(3x+x/2)=0,

    2sin2x.cos2x=0

    adding into parenthesis
    1/2(2sin2x.cos2x)=sin2x

    2sin2x.cos2x=2sin2x

    2sin2x.cos2x - 2sin2x=0

    2sin2x(cos2x-1)=0

    2sin2x=0, cos2x=0

    But using the general equation formula for cos2x=0 gives me x=∏n, when the answer should be x=∏n/2? 2sin2x isn't even close.

    I apologise for my high school level ignorance in advance!
     
  7. Apr 18, 2012 #6
    It looks like you made a mistake in the cosine part, look at the formula again. Also, why did you set the right side equal to zero? It looks like you made some mistakes further down, but you need to restart from an earlier point anyway.

    1/2(2sin2x.cos2x)=sin2x

    does not become

    2sin2x.cos2x=2sin2x

    The 1/2 and the 2 merely cancel.
     
  8. Apr 19, 2012 #7
    Note that 3x+x/2 = 7x/2 not 2x what you should write is (3x +x)/2

    sin3x+sinx=2sin((3x+x)/2)cos(.......) I have left the cos blank as e^(i Pi)+1=0 said you need to check the formula to obtain the correct expression.
     
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