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    Some proof questions:

    1. sin (A+B)/ sin (A-B) = tanA + tanB / (tanA - tanB)

    2. sin (A+B) x sin (A-B) = (sinA)^2 - (sinB)^2

    I just got (sinA)^2 x (cosB)^2 - (cosA)^2 x (sinB)^2 but I don't know how to get rid of the cosines to just leave sin.
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    1. LHS expand out and rearrange then convert

    2. ???

    is it possible to learn the whole c3 in one day
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    \frac {sin(A + B)}{sin(A - B)} = \frac {sinAcosB + sinBcosA}{sinAcosB - sinBcosA} = \frac {(sinAcosB + sinBcosA)^2}{(sinAcosB - sinBcosA)(sinAcosB + sinBcosA)}

    = \frac {sin^2Acos^2B + 2sinAcosB + sin^2Bcos^2A}{sin^2Acos^2B - sin^2Bcos^2A}

    \frac {(sinAcosB)^2 + (cosBsinA)^2 + 2sinAcosB}{(sinAcosB)^2 - (cosAsinB)^2}

    well i regret starting doing this...

    toodles
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    for Q1, just expand RHS, its much easier
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    (Original post by morebritishthanindian)
    1. LHS expand out and rearrange then convert

    2. ???

    is it possible to learn the whole c3 in one day
    no but i managed it with s1
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    (Original post by Pheylan)
    \frac {sin(A + B)}{sin(A - B)} = \frac {sinAcosB + sinBcosA}{sinAcosB - sinBcosA} = \frac {(sinAcosB + sinBcosA)^2}{(sinAcosB - sinBcosA)(sinAcosB + sinBcosA)}

    = \frac {sin^2Acos^2B + 2sinAcosB + sin^2Bcos^2A}{sin^2Acos^2B - sin^2Bcos^2A}


    \frac {(sinAcosB)^2 + (cosBsinA)^2 + 2sinAcosB}{(sinAcosB)^2 - (cosAsinB)^2}

    well i regret starting doing this...

    toodles
    ??????????????:eek: i've got no idea how you got that
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    (Original post by kai4321)
    Some proof questions:

    1. sin (A+B)/ sin (A-B) = tanA + tanB / (tanA - tanB)

    2. sin (A+B) x sin (A-B) = (sinA)^2 - (sinB)^2

    I just got (sinA)^2 x (cosB)^2 - (cosA)^2 x (sinB)^2 but I don't know how to get rid of the cosines to just leave sin.
    are these exam style questions???
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    For the second question, remember that sinA^2+cosA^2=1, substitute both cosines and you'll get the answer.
 
 
 
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