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Math Help - Derive the given identities

  1. #1
    Newbie roflcoptur's Avatar
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    Unhappy Derive the given identities

    I have a very basic knowledge (or almost none at all) about identities and deriving.

    1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
    2. 2csc2beta = secbeta(cscbeta)
    3. tanbeta + cotbeta = 2csc2beta
    4. sin2beta/sinbeta - cos2beta/cosbeta = secbeta
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  2. #2
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    Quote Originally Posted by roflcoptur View Post

    1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
    With this type of problems you need to work with only one side of the equation and apply trig identities until you can find the other side of the equation.

     \frac{\cos(\beta)\tan(\beta) + \sin(\beta)}{ \tan (\beta)}

     \frac{\cos(\beta)\tan(\beta) }{ \tan (\beta)}+ \frac{ \sin(\beta)}{ \tan (\beta)}

     \cos(\beta)+ \frac{ \sin(\beta)}{ \tan (\beta)}

     \cos(\beta)+ \cos(\beta)

     2 \cos(\beta)

    Have a look at this site:

    Table of Trigonometric Identities
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  3. #3
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    Quote Originally Posted by roflcoptur View Post
    I have a very basic knowledge (or almost none at all) about identities and deriving.

    1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
    2. 2csc2beta = secbeta(cscbeta)
    3. tanbeta + cotbeta = 2csc2beta
    4. sin2beta/sinbeta - cos2beta/cosbeta = secbeta
    2. 2\csc{2\beta} = \frac{2}{\sin{2\beta}}

     = \frac{2}{2\sin{\beta}\cos{\beta}}

     = \frac{1}{\sin{\beta}\cos{\beta}}

     = \sec{\beta}\csc{\beta}.
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