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    hey, I'm stuck on these questions, can anyone help?

    firstly:
    the acute angles A and B are such that tan A = 4/3 and tan B = 7
    prove that A + B = 3pi / 4

    I've drawn out triangles and labelled the sides, but I can't think of what else to do...


    Secondly:
    Prove that
    2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

    any suggestions?
    cheers
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    (Original post by kimoni)
    hey, I'm stuck on these questions, can anyone help?

    firstly:
    the acute angles A and B are such that tan A = 4/3 and tan B = 7
    prove that A + B = 3pi / 4

    I've drawn out triangles and labelled the sides, but I can't think of what else to do...


    Secondly:
    Prove that
    2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

    any suggestions?
    cheers
    starting with rhs..
    1+(cos^2P-sin^2P)(cos^2Q-sin^2Q))=1+cos^2Pcos^2Q+sin^2Psi n^2Q-sin^2Pcos^2Q-sin^2Qcos^2P

    =1+cos^2Pcos^2Q+sin^2Psin^2Q+cos ^2Pcos^2Q-cos^2Q+sin^2Qsin^2Q-sin^2Q

    =2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-(cos^2Q+sin^2Q)
    =2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-1
    =2(cos^2Pcos^2Q+sin^2Psin^2Q)
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    (Original post by lgs98jonee)
    starting with rhs..
    1+(cos^2P-sin^2P)(cos^2Q-sin^2Q))=1+cos^2Pcos^2Q+sin^2Psi n^2Q-sin^2Pcos^2Q-sin^2Qcos^2P

    =1+cos^2Pcos^2Q+sin^2Psin^2Q+cos ^2Pcos^2Q-cos^2Q+sin^2Qsin^2Q-sin^2Q

    =2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-(cos^2Q+sin^2Q)
    =2(cos^2Pcos^2Q+sin^2Psin^2Q)+1-1
    =2(cos^2Pcos^2Q+sin^2Psin^2Q)
    Ah, cheers.. I didn't know you could start from the RHS, is it acceptable?
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    (Original post by kimoni)
    hey, I'm stuck on these questions, can anyone help?

    firstly:
    the acute angles A and B are such that tan A = 4/3 and tan B = 7
    prove that A + B = 3pi / 4

    I've drawn out triangles and labelled the sides, but I can't think of what else to do...
    Think about what tan(A+B) is...

    And its fine to start with RHS because the expression is tautologous -it is identical for all x
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    (Original post by It'sPhil...)
    tautologous
    u cant make up words
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    (Original post by lgs98jonee)
    u cant make up words
    I know. I didn't
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    (Original post by It'sPhil...)
    I know. I didn't
    hehe

    hey heres a question for ya.

    when youve got a fraction in a fraction, how do you sort it out?

    i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
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    (Original post by kikzen)
    hehe

    hey heres a question for ya.

    when youve got a fraction in a fraction, how do you sort it out?

    i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
    first make the 1-(1/x) a single fraction:
    (x-1)/x
    then multiply the 'x' to the numerator
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    (Original post by kikzen)
    hehe

    hey heres a question for ya.

    when youve got a fraction in a fraction, how do you sort it out?

    i mean things like 1-(1/x) on the bottom of the fraction... you can times the x up, but where else do you have to times it? the top too?
    What i do is times every single term by the fraction bit, including the top ones...

    Eg 1/(3x + 1/sinx) ... times everything by sinx to get sinx/(3x.sinx + 1).
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    (Original post by kimoni)
    hey, I'm stuck on these questions, can anyone help?

    firstly:
    the acute angles A and B are such that tan A = 4/3 and tan B = 7
    prove that A + B = 3pi / 4

    I've drawn out triangles and labelled the sides, but I can't think of what else to do...


    Secondly:
    Prove that
    2(sin^2(P)*sin^2(Q)) + 2(cos^2(P)*cos^2(Q)) = 1 + cos2P cos2Q

    any suggestions?
    cheers
    tanA+tanB=tan(A+B)(1-tanAtanB)
    4/3+7=tan(A+B)(1-4/3.7)
    tan(A+B)=-1
    A+B=arctan(-1)
    A+B=3pi/4.
 
 
 
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