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Math Help - Trigonmetric Identity - Tan Theta

  1. #1
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    Trigonmetric Identity - Tan Theta

    What you have to multiply for addition/subtraction identity, I was thinking of getting cosB or sinB out.

    tan theta = cos (A+B) / sin (A+B)
    = cosAcosB - sinAsinB / sinAcosB + cosAsinB
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  2. #2
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    Quote Originally Posted by Quan View Post
    What you have to multiply for addition/subtraction identity, I was thinking of getting cosB or sinB out.

    tan theta = cos (A+B) / sin (A+B)
    = cosAcosB - sinAsinB / sinAcosB + cosAsinB
    First of all, use parenthesis.
    tan(\theta) = \frac{cos(A + B)}{sin(A + B)} = \frac{cos(A)cos(B) - sin(A)sin(B)}{sin(A)cos(B) + cos(A)sin(B)}

    not
    tan(\theta) = \frac{cos(A + B)}{sin(A + B)} = cos(A)cos(B) - \frac{sin(A)sin(B)}{sin(A)cos(B)} + cos(A)sin(B)

    Second, what are you asking for? Your question is very vague.

    -Dan
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    Quote Originally Posted by topsquark View Post
    First of all, use parenthesis.
    tan(\theta) = \frac{cos(A + B)}{sin(A + B)} = \frac{cos(A)cos(B) - sin(A)sin(B)}{sin(A)cos(B) + cos(A)sin(B)}

    not
    tan(\theta) = \frac{cos(A + B)}{sin(A + B)} = cos(A)cos(B) - \frac{sin(A)sin(B)}{sin(A)cos(B)} + cos(A)sin(B)

    Second, what are you asking for? Your question is very vague.

    -Dan
    Sorry, I'm looking for the addition / subtraction Identities for Tan Theta
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quan View Post
    Sorry, I'm looking for the addition / subtraction Identities for Tan Theta
    By the way, I missed this in the previous post:
    tan(\theta) = \frac{sin(\theta)}{cos(\theta)}
    not the other way around.

    tan(A + B) = \frac{sin(A + B)}{cos(A + B)} = \frac{sin(A)cos(B) + cos(A)sin(B)}{cos(A)cos(B) - sin(A)sin(B)}

    Now divide the numerator and denominator by cos(A)cos(B):
    tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}

    -Dan
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