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Math Help - Trig Identity?

  1. #1
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    Trig Identity?

    I recently read somewhere that sin(\theta)=\frac{y}{\sqrt{y^2+x^2}}, but I cannot seem to find this identity anywhere on the internet. Does anyone know what this is called, or where I can find the same identities (with x's and y's) for the other trig functions?
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    You should know that \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2 \implies r = \sqrt{x^2 + y^2}.


    So \displaystyle r\sin{\theta} = y

    \displaystyle \sin{\theta} = \frac{y}{r}

    \displaystyle \sin{\theta} = \frac{y}{\sqrt{x^2 + y^2}}.
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  3. #3
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    is it accurate to say that:

    cos(\theta)=\frac{x}{\sqrt{y^2+x^2}}?
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    Quote Originally Posted by bobbooey View Post
    is it accurate to say that:

    cos(\theta)=\frac{x}{\sqrt{y^2+x^2}}?
    Yes
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  5. #5
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    Hello, bobbooey!

    \text{I recently read somewhere that: }\:\sin(\theta)\:=\:\dfrac{y}{\sqrt{x^2+y^2}}

    \text{Does anyone know the identities for the other trig functions?}

    Take a look at where that equation comes from . . .

    Code:
            |
            |           *
            |         * |
            |    h  *   |
            |     *     | y
            |   *       |
            | * @       |
        - - * - - - - - + - - 
            |     x

    We have angle \,\theta in a right triangle.

    The adjacent side is \,x; the opposite side is \,y.

    Using Pythagorus, we find that: . h \:=\:\sqrt{x^2+y^2}


    Therefore, we have:

    . . \begin{array}{cccccc}<br />
\sin\theta &=& \dfrac{opp}{hyp} &=& \dfrac{y}{\sqrt{x^2+y^2}} \\ \\[-3mm]<br />
\cos\theta &=& \dfrac{adj}{hyp} &=& \dfrac{x}{\sqrt{x^2+y^2}} \\ \\[-3mm]<br />
\tan\theta &=& \dfrac{opp}{adj} &=& \dfrac{y}{x} \\ \\[-3mm]<br />
\cot\theta &=& \dfrac{adj}{opp} &=& \dfrac{x}{y} \\ \\[-3mm]<br />
\sec\theta &=& \dfrac{hyp}{adj} &=& \dfrac{\sqrt{x^2+y^2}}{x} \\ \\[-3mm]<br />
\csc\theta &=& \dfrac{hyp}{opp} &=& \dfrac{\sqrt{x^2+y^2}}{y} \end{array}

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