# General trig question

• Jan 9th 2013, 07:21 AM
kinhew93
General trig question
When cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

But when written as cos(x)[1 + tan(x)] = 0 cos(x) can be 0?

Could someone please try to explain this to me :)
• Jan 9th 2013, 07:42 AM
ebaines
Re: General trig question
Quote:

Originally Posted by kinhew93
When cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

But when written as cos(x)[1 + tan(x)] = 0 cos(x) can be 0?

Could someone please try to explain this to me :)

If $\displaystyle \cos (x) = 0$ then $\displaystyle \tan(x)=\infty$, and the product of $\displaystyle 0 \times \infty$ is undefined. So no: if $\displaystyle \cos(x)=0$, then $\displaystyle \cos(x)[ 1 + \tan(x)] \ne 0$.
• Jan 9th 2013, 08:20 AM
Soroban
Re: General trig question
Hello, kinhew93!

Quote:

$\displaystyle \text{When }\,\cos x + \sin x \:=\:0,\,\text{ I can see that }\cos x \ne 0$

$\displaystyle \text{But when written as: }\,\cos x(1 + \tan x) \:=\:0,\: \cos x\text{ can be 0.}$

$\displaystyle \text{Could someone please try to explain this to me?}$

We have: .$\displaystyle \sin x \:=\:-\cos x$

. . . . . . . . $\displaystyle \frac{\sin x}{\cos x} \:=\:-1$ . . . You are right: $\displaystyle \cos x \ne 0$

. . . . . . . . $\displaystyle \tan x \:=\:-1$

Therefore: n . $\displaystyle x \:=\:-\tfrac{\pi}{4} + \pi n$