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Math Help - Showing Trigonometric Identities

  1. #1
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    Showing Trigonometric Identities

    Hi all,

    Just registered with this excellent-looking site. Hoping to be able to help others with their problems as well as receiving help myself with some trickier problems!

    I'm going to start off with a puzzle that I have been asked recently...

    Show that sin^2(x) = 1 - cos^2(x)

    I understand that this is a half-angle formula, and know how to prove it using rearrangements of the double angle formulas/compound angles. However, I am unsure how to show that the LHS = RHS using the basic trigonometric identities, especially cos^2(x) + sin^2(x) = 1.

    Thanks to all in advance.

    Lee
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  2. #2
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    Hello lee!
    Not sure what your getting at Lee really! The identity you speak of is on of the basic identies that is used to solve many problems. I can explain its origins if that helps???? If not ignore!!!

    Imagine the graph of a circle with the origin as its center and its radius of 1 unit. Now draw a Right angled triangle with the radius as its hypoyonuse which originates at the origin and obviously meets the circumference. So if we now take the angle made between the hypotonuse and the base of the triangle as X we can work out the two unknown side lengths as :

    Base length: cos x = base/1
    cos x = base

    Opposite length: sin x = opposite/1
    sin x = opposite

    so now we use these values in pythagorous theorem thus...

    1^2 = sin^2x + cos^2x

    So the proof on the equation i suppose is just pythagorous!

    Chappers
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  3. #3
    Junior Member
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    Jul 2008
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    Smile

    Hi

    By using the pythogras theorem

    sum of the square of the two sides of a triangle is equal to hypotenuse square side

    that is

    (opposite side)^2+(adjacent side)^2=(hypotenuse side)^2
    dividing by (hypotenuse side)^2 on both the sides, we get

    (opp side/hyp side)^2 +(adj side/hyp side)^2=1

    sin^2x+cos^2x=1
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  4. #4
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    Jul 2008
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    Thanks both for the reply. The reason why you're not sure what I'm getting at gtbiyb is because I entered the expression incorrectly!

    What I am trying to show is that:

    sin^2(x) = (1-cos(2x))/2

    The expression I wrote originally, sin^2(x) = 1 - cos^2(x), was the first step I used in trying to show that the LHS = RHS.

    So far I have:

    sin^2(x) = 1 - cos^2(x) ... since cos^2(x) + sin^2(x) = 1

    The trouble I am having is converting the 1 - cos^2(x) into (1-cos(2x))/2 using basic trigonometric identities.

    I appreciate any further help.

    Thanks.
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  5. #5
    Junior Member
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    Jul 2008
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    Hi,

    To show that 1-cos^2x=(1-cos2x)/2

    use the formula for cos^2x=(Cos2x+1)/2

    LHS
    =1-cos^2x
    =1-[(1+cos2x)/2]
    =(2-1-cos2x)/2
    =(1-cos2x)/2 RHS

    LHS=RHS
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