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Math Help - trig identities, proof

  1. #1
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    trig identities, proof

    prove these identities;

    A)  \frac{1-cos\theta}{sin\theta} =  \frac{sin\theta}{1+cos\theta}


    b)  (\frac{1}{cos\theta} + tan\theta) (\frac{1}{cos\theta} - tan\theta) = 1

    is this working correct for the second one;

     y = \frac{1}{cos\theta}

     (y+tan\theta) (y-tan\theta)

     y^2 - tan^2\theta = 1

     y^2 - \frac{sin^2\theta}{cos^2\theta}

    where would I go from here?
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  2. #2
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    Quote Originally Posted by Tweety View Post
    prove these identities;

    A)  \frac{1-cos\theta}{sin\theta} =  \frac{sin\theta}{1+cos\theta}


    b)  (\frac{1}{cos\theta} + tan\theta) (\frac{1}{cos\theta} - tan\theta) = 1

    is this working correct for the second one;

     y = \frac{1}{cos\theta}

     (y+tan\theta) (y-tan\theta)

     y^2 - tan^2\theta = 1

     y^2 - \frac{sin^2\theta}{cos^2\theta}

    where would I go from here?
    Remember what you made y equal to and put it back in:

    y^2 = \frac{1}{cos^2\theta}

    This gives you

    \frac{1-sin^2\theta}{cos^2\theta} = \frac{cos^2\theta}{cos^2\theta} = 1 = RHS


    For number 1 get the same denominator by cross multiplying:

     \frac{1-cos\theta}{sin\theta} =  \frac{sin\theta}{1+cos\theta}

    \frac{(1-cos\theta)(1+cos\theta)}{sin\theta(1+cos\theta)} = \frac{sin^2\theta}{sin\theta(1+cos\theta)}

    The denominators cancel and the LHS is the difference of two squares:

    1-cos^2\theta = sin^2\theta

    sin^2\theta = sin^2\theta
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  3. #3
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    How does getting this expression sin^2\theta = sin^2\theta prove that  \frac{1-cos\theta}{sin\theta} =  \frac{sin\theta}{1+cos\theta} ?

    Are you not meant to make the first expression look like the second one?

    Thanks,
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  4. #4
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    Talking

    Follow the listed steps: Take that second equality, cross-multiply, and simplify using the listed identities. What do you get?
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  5. #5
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    Hello, Tweety!

    Prove these identities:

    . . (a)\;\;\frac{1-\cos\theta}{\sin\theta} \:=\: \frac{\sin\theta}{1+\cos\theta}

    Multiply the left side by \frac{1+\cos\theta}{1+\cos\theta}

    \frac{1-\cos\theta}{\sin\theta}\cdot\frac{1+\cos\theta}{1+  \cos\theta} \:=\:\frac{1-\cos^2\!\theta}{\sin\theta(1 + \cos\theta)} \:=\:\frac{\sin^2\!\theta}{\sin\theta(1 + \cos\theta)} \;=\;\frac{\sin\theta}{1 + \cos\theta}



    (b)\;\;\left(\frac{1}{\cos\theta} + \tan\theta\right)\left(\frac{1}{\cos\theta} - \tan\theta\right) \:=\: 1

    The left side is: . (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) \:=\:\sec^2\!\theta - \tan^2\!\theta \;=\;1

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  6. #6
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    Quote Originally Posted by Tweety View Post
    prove these identities;

    A)  \frac{1-cos\theta}{sin\theta} =  \frac{sin\theta}{1+cos\theta}


    b)  (\frac{1}{cos\theta} + tan\theta) (\frac{1}{cos\theta} - tan\theta) = 1

    is this working correct for the second one;

     y = \frac{1}{cos\theta}

     (y+tan\theta) (y-tan\theta)

     y^2 - tan^2\theta = 1

     y^2 - \frac{sin^2\theta}{cos^2\theta}

    where would I go from here?
    1.You multiply on the diagonal and get:
    sin^2x=(1-cosx)(1+cosx) \Leftrightarrow sin^2x=1-cos^2x \Leftrightarrow sin^2x+cos^2x=1

    2.You have tanx=\frac{sinx}{cosx}
    So
    <br />
(\frac{1}{cosx}+tanx)(\frac{1}{cosx}-tanx)=1 <br />

    <br />
(\frac{1}{cosx}+\frac{sinx}{cosx})(\frac{1}{cosx}-\frac{sinx}{cosx})=1

    <br />
\frac{(1+sinx)(1-sinx)}{cos^2x}=1

     \frac{1-sin^2x}{cos^2x}=1

    <br />
\frac{cos^2x}{cos^2x}=1<br />
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