Trig Identity?

**bobbooey** · December 13th 2010, 06:31 PM

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?

**Prove It** · December 13th 2010, 06:33 PM

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

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

$\displaystyle \sin{\theta} = \frac{y}{\sqrt{x^2 + y^2}}$ .

**bobbooey** · December 13th 2010, 06:51 PM

is it accurate to say that:

$cos(\theta)=\frac{x}{\sqrt{y^2+x^2}}$ ?

**Prove It** · December 13th 2010, 06:54 PM

Originally Posted by bobbooey

is it accurate to say that:

$cos(\theta)=\frac{x}{\sqrt{y^2+x^2}}$ ?

Yes

**Soroban** · December 13th 2010, 08:24 PM

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:

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

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