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Math Help - Compound Angle Identity

  1. #1
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    Compound Angle Identity

    How would I prove these identities?

    1) sin(A+B)*sin(A-B)=(sin A)^2*(sin B)^2

    I've gotten to expanding it to
    (sin(A)cos(B)+sin(B)cos(A))*(sin(A)cos(B)-sin(B)cos(A))
    then

    (sin(A)cos(B))^2-(sin(B)cos(A))^2, but I'm not sure what to do next

    2) (cos W-sin 2W)/(cos 2W+sin W-1)=cot W
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  2. #2
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    GHello, DemonX01!

    Please check the first one for typos . . .


    2)\;\frac{\cos x-\sin2x}{\cos2x+\sin x-1} \;=\;\cot x

    Note: If x = \tfrac{\pi}{6}, the denominator is: . \cos\tfrac{\pi}{3} + \sin\tfrac{\pi}{6} - 1 \:=\:\tfrac{1}{2} + \tfrac{1}{2}-1 \;=\;0

    Hence: . x \,\neq\,\tfrac{\pi}{6}

    . . That is: . \sin x \,\neq\, \tfrac{1}{2} \quad\Rightarrow\quad 2\sin x \,\neq\, 1 \quad\Rightarrow\quad 1 - 2\sin x\:\neq\:0 . .
    We can cancel this



    We have: . \frac{\cos x - \sin2x}{\cos2x + \sin x - 1} \;=\;\frac{\cos x - 2\sin x\cos x}{(1-2\sin^2\!x) + \sin x - 1} \;=\;\frac{\cos x - 2\sin x\cos x}{\sin x - 2\sin^2\!x}

    Factor: . \frac{\cos x(1 - 2\sin x)}{\sin x(1 - 2\sin x)}


    Reduce: . \frac{\cos x}{\sin x} \;=\;\cot x

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  3. #3
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    Thanks for the help on the second one. But as far as I see, there aren't any typos in the first one that I notice. Is there something wrong?
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  4. #4
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    Hello again, DemonX01!

    1)\;\;\sin(A+B)\cdot\sin(A-B)\:=\:{\color{blue}\sin^2\!A \cdot\sin^2\!B} .??
    Your work is correct!

    \bigg[\sin A\cos B+ \cos A\sin B\bigg]\cdot\bigg[\sin A\cos B -\cos A\sin B\bigg]

    . . . =\qquad\sin^2\!A\cos^2\!B \quad - \quad \cos^2\!A\sin^2\!B
    . . . =\;\sin^2\!A\overbrace{\left(1-\sin^2\!B\right)} - \overbrace{\left(1 - \sin^2\!A\right)}\sin^2\!B

    . . . =\; \sin^2\!A - \sin^2\!A\sin^2\!B - \sin^2\!B + \sin^2\!A\sin^2\!B

    . . . =\;\sin^2\!A - \sin^2\!B

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  5. #5
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    Thanks for the help! I just keep on forgetting about the basic identities
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