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Math Help - tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

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    tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

    tanA=12/5 and A=6B then find 12cosec2B + 5sec2B
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    Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

    Okay, do it! Start with a Right Triangle. Label A and B. Lable the legs so that tan(A) = 12/5. Determine the hypotenuse and you are well on your way.
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    Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

    Quote Originally Posted by TKHunny View Post
    Okay, do it! Start with a Right Triangle. Label A and B. Lable the legs so that tan(A) = 12/5. Determine the hypotenuse and you are well on your way.
    but that will be tan A and lets say ill get the value of tan6B and then how will i get tan 2B
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    Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

    If A= 6B then B= A/6 so 2B= A/3.

    Hopefully, you know that sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y) and that cos(x+y)= cos(x)cos(y)- sin(x)sin(y). Taking x= y, sin(2x)= 2sin(x)cos(x) and cos(2x)= cos^2(x)- sin^2(x). Then sin(3x)= sin(2x+ x)= sin(2x)cos(x)+ cos(2x)sin(x)= 2sin(x)cos^2(x)+ (cos^2(x)- sin^2(x))sin(x)= 3sin(x)cos^2(x)- sin^3(x).
    cos(3x)= cos(2x+ x)= cos(2x)cos(x)- sin(2x)sin(x)= (cos^2(x)- sin^2(x))cos(x)- 2sin^2(x)cos(x)= cos^3(x)- sin^2(x)cos(x).

    If x= A/3, then
    sin(3x)= sin(A)= 3sin(A/3)cos^2(A/3)- sin^3(A/3)= 3sin(B)cos^2(B)- sin^3(B)
    and
    cos(3x)= cos(A)= cos^3(A/3)- sin^2(A/3)cos(A/3)= cos^3(B)- sin^2(B)cos(B).

    From tan A= 12/5, you can solve for sin(A) and cos(A). Then you have two equations to solve for sin(B) and cos(B).
    Last edited by HallsofIvy; July 19th 2011 at 08:00 AM.
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