# Trig Identity?

• Dec 13th 2010, 06:31 PM
bobbooey
Trig Identity?
I recently read somewhere that $\displaystyle sin(\theta)=\frac{y}{\sqrt{y^2+x^2}}$, 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?
• Dec 13th 2010, 06:33 PM
Prove It
You should know that $\displaystyle \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2 \implies r = \sqrt{x^2 + y^2}$.

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

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

$\displaystyle \displaystyle \sin{\theta} = \frac{y}{\sqrt{x^2 + y^2}}$.
• Dec 13th 2010, 06:51 PM
bobbooey
is it accurate to say that:

$\displaystyle cos(\theta)=\frac{x}{\sqrt{y^2+x^2}}$?
• Dec 13th 2010, 06:54 PM
Prove It
Quote:

Originally Posted by bobbooey
is it accurate to say that:

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

Yes
• Dec 13th 2010, 08:24 PM
Soroban
Hello, bobbooey!

Quote:

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

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

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

Code:

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

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

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

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

Therefore, we have:

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