# Thread: Rearranging an equation using a simple trigonometric identity

1. ## Rearranging an equation using a simple trigonometric identity

Hi all,

I have a mental block on a certain very simple trigonometry problem. I know I will slap my forehead when I see the solution. I therefore ask for your indulgence and thank, in advance, any responders. Here is the question and my attempt:

Using $\displaystyle \tan \theta \equiv \frac{\sin \theta}{\cos \theta}$, show that the equation $\displaystyle \tan \theta = \sin \theta$ can be written as $\displaystyle \sin \theta (\cos \theta - 1) = 0$ provided that $\displaystyle \cos \theta \neq 0$.

Here is what I have done already:

$\displaystyle \tan \theta = \sin \theta$

$\displaystyle \sin \theta = \frac{\sin \theta}{\cos \theta}$

$\displaystyle \sin \theta = \sin \theta \cos \theta$

$\displaystyle \cos \theta = 1$

I am not at all sure that I am on the right track. Any help would be sincerely appreciated.

Regards,

Evanator

2. Originally Posted by evanator
Hi all,

I have a mental block on a certain very simple trigonometry problem. I know I will slap my forehead when I see the solution. I therefore ask for your indulgence and thank, in advance, any responders. Here is the question and my attempt:

Using $\displaystyle \tan \theta \equiv \frac{\sin \theta}{\cos \theta}$, show that the equation $\displaystyle \tan \theta = \sin \theta$ can be written as $\displaystyle \sin \theta (\cos \theta - 1) = 0$ provided that $\displaystyle \cos \theta \neq 0$.

Here is what I have done already:

$\displaystyle \tan \theta = \sin \theta$

$\displaystyle \sin \theta = \frac{\sin \theta}{\cos \theta}$

$\displaystyle \sin \theta = \sin \theta \cos \theta$

starting from the above step ...

$\displaystyle 0 = \sin{t}\cos{t} - \sin{t}$

$\displaystyle 0 = \sin{t}(\cos{t} - 1)$

3. I was right. It was a doh! moment. Thank you very much, skeeter.