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Homework Help: Simplifying Trig Products

  1. Oct 30, 2007 #1
    [SOLVED] Simplifying Trig Products

    1. The problem statement, all variables and given/known data
    Express the following as a product and simplify.
    sin60° + sin20°

    2. Relevant equations

    3. The attempt at a solution

    I don't understand what the question is trying to say. For example, do I convert sin60° and then add. I don't understand. Someone please help.
  2. jcsd
  3. Oct 30, 2007 #2
  4. Oct 30, 2007 #3
    So, [tex]\sqrt{3}/2[/tex] + sin20°
    = 1.208

    Could this be right?
    Last edited: Oct 30, 2007
  5. Oct 30, 2007 #4
    Hi rum2563,
    I dont think you are required to solve it, you just ned to rewrite it in a different form.
    You will need to use a trig identity which converts the sum to a product.
    Moridin has pointed you in the right direction with the hint: "sin a + sin b = some product".
    Look on the webpage Moridin has provided a link to and the equation you need is located under Sum-to-Product Formulas.
  6. Oct 30, 2007 #5
    Thanks Evalesco.

    So, here it is:

    sin60° + sin20° = 2sin(60+20/2)cos(60+20/2)

    Is this right?

    Also, because of the fact that you pointed this to me, there is another question I have.

    The question states:
    Using the method developed in Example 3 (my note: we don't have the book so we don't know what this method is) of this section, prove each of the following Transformation Formulas.

    sinx - siny = 2cos(x+y/2)sin(x-y/2)

    Could someone please help me with this? Also, thanks Evalesco and Moridin for helping me out.
  7. Oct 30, 2007 #6
    Not quite; apply the correct formula under the headline "Sum-to-Product Formulas" and do not forget the signs or what should be divided with two.
  8. Oct 30, 2007 #7
    http://hh4.hollandhall.org/kluitwieler/pages/Advanced_Trig/Trig%20Frog%20Homepage/sumtoproduct.htm [Broken]

    About 10 seconds ago, I looked at the above webpage.

    Here is what I came up with now:

    sin60° + sin20° = 2sin40°cos20°

    I hope finally it is correct.

    Also, can you please help me with my second question which I posted in the last post? Thanks.
    Last edited by a moderator: May 3, 2017
  9. Oct 30, 2007 #8
    Yep it's correct.
    Did you mean sinx - siny = 2cos((x+y)/2)sin((x-y)/2)?

    I will start you out on one way of showing the above

    Start out with sinx - siny

    [tex] Then \ let \ x=\frac{x+y}{2}+\frac{x-y}{2} \ and \ let \ y=\frac{x+y}{2}-\frac{x-y}{2}[/tex]

    You will find the identities sin(a+b) = sin(a)cos(b)+cos(a)sin(b) and sin(a-b) = sin(a)cos(b)-cos(a)sin(b) will come in handy for the next step.

    Let me know how you get on.
  10. Oct 31, 2007 #9
    Yes, I finally got it.

    sin60° + sin20° = 2sin((x+y)/2)cos((x-y)/2)

    A + B = 60°
    A - B = 20°
    ---------- +
    2A = 80°
    A = 40°

    A + B = 60°
    A - B = 20°
    ---------- -
    2B = 40°
    B = 20°

    sin60° + sin20° = 2sin(40°)cos(20°)

    Yes, I finally got it. But could there have been an easier way? Or is this good enough? I tried my best anyways.

    Thanks to Evalesco and Moridin for helping me out.
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