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Math Help - Verify My Identity - Impossible?

  1. #1
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    Verify My Identity - Impossible?

    Our teacher gave us this problem to work on this week, but I think either he or someone else copied it down wrong. I have tried everything I can think of; see if any of you can get it:

    1-sin(t) / cos(t) = cos(t) / 1-sin(t)
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  2. #2
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    Quote Originally Posted by Gambit View Post
    Our teacher gave us this problem to work on this week, but I think either he or someone else copied it down wrong. I have tried everything I can think of; see if any of you can get it:

    1-sin(t) / cos(t) = cos(t) / 1-sin(t)
    Let's see: (And please use parenthesis!)
    \frac{1 - sin(t)}{cos(t)}

    = \frac{1 - sin(t)}{cos(t)} \cdot \frac{1 + sin(t)}{1 + sin(t)}

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

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

    = \frac{cos(t)}{1 + sin(t)}

    You are right. It was copied wrong.

    -Dan
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  3. #3
    Oli
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    This is false:
    Let t=Pi/3 and stick the numbers in and you will see.
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  4. #4
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    Hello, Gambit!

    The classic identity is: . \frac{1-\sin t}{\cos t} \;=\;\frac{\cos t}{1+ \sin t} . . . or one its variations.

    . . The two binomials must have opposite signs.



    I've created dozens of identities for my student's exams.


    Start with \sin^2\!x + \cos^2\!x \:=\:1 . . .

    . . 1-\sin x)" alt="\cos^2\!x\:=\:1-\sin^2\!x \quad \Rightarrow\quad (\cos x)(\cos x) \:=\1-\sin x)" /> (1 + \sin x) \quad \Rightarrow \quad \frac{\cos x}{1-\sin x} \:=\:\frac{1+\sin x}{\cos x}

    . . 1-\cos x)" alt="\sin^2\!x\:=\:1-\cos^2\!x \quad\Rightarrow\quad (\sin x)(\sin x) \:=\1-\cos x)" /> (1 + \cos x) \quad\Rightarrow\quad \frac{\sin x}{1+\cos x} \:=\:\frac{1-\cos x}{\sin x}

    . . . See?

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