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Addition and subtraction formula, proving identities

  1. Mar 26, 2008 #1
    The question states, "Prove the identity."
    (1) cos(x + y) cos(x-y) = cos(^2)x – sin(^2)y.
    Should i start off using the addition and subtraction formulas for the LHS, and breaking down the perfect square for the RHS? If not or if so, how would I go about solving this problem? Step by step.

    (2) sin(x + y + z) = sinx cosy cosz + cosx siny cosz + cosx cosy sinz - sinx siny sinz.
    I started this off using the addition formula for the LHS, but I ended up splitting the LHS into two separate identities: sin(x + y) sin(y + z). How do I approach this problem?
  2. jcsd
  3. Mar 26, 2008 #2


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    Hi cheab14! :smile:

    No … don't start on both sides and try and make them meet in the middle!

    You're not building a tunnel! :frown:

    Just stick to one side, and try to make it equal the other.

    In this case, do the LHS, exactly the way you suggested. :smile:
    Show us how you did that. :confused:
  4. Mar 26, 2008 #3
    hello tiny-tim!!

    for (1) i worked out the LHS like so:
    (cosxcosy - sinxsiny)(cosxcosy + sinxsiny)
    which ended up being: cos(^2)x cos(^2)y - sin(^2)x sin(^2)y. And then I'm stuck.

    for (2) I'm just lost:confused:
    -I have: sinxcosy + cosxsiny + sinycosz + cosysinz for the LHS which makes no sense to me:frown:
  5. Mar 26, 2008 #4


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    Hi cheab14! :smile:

    Well, you know the answer has only cos(^2)x and sin(^2)y.

    So just rearrange the sin(^2)x and cos(^2)y in terms of cos(^2)x and sin(^2)y ! :smile:

    (Alternatively, you could have used the rule cosAcosB = (cos(A+B) + cos(A-B))/2.)
    How did you get that? :confused:

    Just use sin(A+B) = … , with A = x+y and B = z. :smile:
  6. Mar 26, 2008 #5
    ok thank u much!!
  7. Mar 26, 2008 #6


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    A specific graph can be created from which the addition formula can be very understandably derived. Try an internet search since I cannot remember what books, or site showed it. You might also find this in an old Calculus book (but not using any calculus) written by Anton.
  8. Mar 26, 2008 #7


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    … all together now …

    You can also sing it to the tune of The Battle Hymn of the Republic:

    Sin A plus B equals sin A cos B plus cos A sin B

    Cos A plus B equals cos A cos B minus sin A sin B

    Sin A minus B equals sin A cos B minus cos A sin B

    CosAminusB equals cos A cos B plus sin A sin B! :smile:

    :smile: It's your patriotic duty to learn it! :smile:
    © tiny-tim ...
  9. Mar 26, 2008 #8
    lol ok...i needed a way to remember those formulas...but i have two other questions: one involving proving identities...n the other involving expressions:
    (1)prove the identity:
    tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y) tan(y-z)tan(z-x)
    for this i started using the subraction formula on the LHS, but then that didn't seem to get me any closer. Then I decided to use tan in terms of sin and cos, now I don't what to do.

    (2)write the expression in terms of sine only:
    (the square root of 3)(sinx) + cosx.
    At first I said that the square root of 3 was tan(pi/3) and then broke tan down into sin and cos components still multiplying that function by sinx; then i used the double angle formula for cosx which gave me the square root of of:(1-sin(^2)x) for cosx. Then I didn't know where to go from there.
  10. Mar 27, 2008 #9


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    Hi cheab14! :smile:

    (1) To save on typing, I'll put tanx = X, tany = Y, tanz = Z.

    Then LHS x (1+XY)(1+YZ)(1+ZX)
    = (X-Y)(1+YZ)(1+ZX)
    + (Y-Z)(1+XY)(1+ZX)
    + (Z-X)(1+XY)(1+YZ);

    you need to show that the only items which don't cancel are those with three letters;
    well, the 1s obviously cancel;
    and then … ? :smile:

    (2) When it says " in terms of sine only", that means sine and numbers! - so you don't have to convert the √3.
    So you had the answer already: √3.sinx + √(1 - sin^2x). :smile:
  11. Mar 27, 2008 #10
    ohk i'll keep (1) in mind when I work it out again thank u! and number (2)-couldn't believe it was that simple..thank u thank u thank u!
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