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Math Help - Proving Trig Identity: Sine Difference

  1. #1
    Newbie
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    Mar 2007
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    Proving Trig Identity: Sine Difference

    In this figure, the triangle has hypotenuse 1.

    I need to use it to show that:

    <br />
sin(A - B) = sinAcosB - sinBcosA<br />

    The hint is :
    <br />
sin(A - B) = \frac{X}{h}<br />
    Find other expressions for X and h and substitute them into this equation.

    I know this is easy. But I'm just stuck. I've tried working out other side lengths and looking at other trig identities but I'm just at a brick wall without a clue why.

    Thanks for any help.
    Attached Thumbnails Attached Thumbnails Proving Trig Identity: Sine Difference-tri_di.gif  
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  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
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    288

    Answer

    Greetings. You don't need any other trig identities. Everything is on the diagram

    Step 1. Fill in the angles, and you'll see that the 3rd from left triangle and the large triangle are similar.

    Step 2. Find the scale factor.
    In this case, the large triangle has a hypotenuse of 1 and the small triangle has a hypotenuse of sinA-hsinB
    Hence scale factor(Q): Q=\frac{1}{sinA-hsinB}

    Step 3. Apply this to the other side of the triangle, opposite the angle of (90-A).
    QX=cosA
    Hence: X=cosAsinA-hcosAsinB

    Put into
    And you get: sin(A-B)=\frac{cosAsinA-hcosAsinB}{h}

    Simplify, then look at your triangle and realize that
    cosB=\frac{cosA}{h}

    Substitute and you arrive at the formula
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  3. #3
    Newbie
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    Excellent solution I-think!

    Cleared that right up.

    Thank you so much.
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