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Thread: Verifying that this is an identity?

  1. August 5th 2011, 12:58 PM #1
    explodingtoenails
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    Verifying that this is an identity?

    Trig and I are not very good friends...

    \displaystyle \frac{1 + tan^2x}{csc^2x} = tan^2x

    This is what I tried:

    \displaystyle \frac{1 + (sin^2x/cos^2x)}{(1/sin^2x)} = tan^2x

    \displaystyle \frac{cos^2x + sin^2x}{cos^2x} * \frac{sin^2x}{1} = tan^2x

    (sin^2x)(sin^2x) = tan^2x

    I don't know why it's not equaling up. D:
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  2. August 5th 2011, 01:03 PM #2
    TheChaz
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    Re: Verifying that this is an identity?

    Quote Originally Posted by explodingtoenails View Post
    Trig and I are not very good friends...

    \displaystyle \frac{1 + tan^2x}{csc^2x} = tan^2x

    This is what I tried:

    \displaystyle \frac{1 + (sin^2x/cos^2x)}{(1/sin^2x)} = tan^2x

    \displaystyle \frac{cos^2x + sin^2x}{cos^2x} * \frac{sin^2x}{1} = tan^2x

    ...

    I don't know why it's not equaling up. D:
    From here, the top left is ONE.
    Then you have (sin/cos)^2
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  3. August 5th 2011, 01:06 PM #3
    explodingtoenails
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    Re: Verifying that this is an identity?

    Thank you so much dude. I didn't know that cos^2x + sin^2x = 1.
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  4. August 5th 2011, 01:42 PM #4
    TheChaz
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    Re: Verifying that this is an identity?

    Quote Originally Posted by explodingtoenails View Post
    Thank you so much dude. I didn't know that cos^2(x) + sin^2(x) = 1.
    That's kind of important! From it, many (most?) identities are derived.
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  5. August 15th 2011, 07:55 PM #5
    Drexel28
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    Re: Verifying that this is an identity?

    Quote Originally Posted by TheChaz View Post
    That's kind of important! From it, many (most?) identities are derived.
    The most important identity really. By this I mean that as rings \mathbb{R}[x,y]/(x^2-y^2-1)\cong \mathbb{R}[\cos(\theta),\sin(\theta)]. In other words, if you take the ring \mathbb{R}[x,y] and impose the conditions that x^2+y^2=1 then you really just have the ring of trigonometric polynomials \mathbb{R}[\cos(\theta),\sin(\theta)].
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