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Thread: General trig question

  1. #1
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    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
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: General trig question

    Quote Originally Posted by kinhew93 View Post
    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 $\displaystyle \cos(x)[ 1 + \tan(x)] \ne 0$.
    Last edited by ebaines; Jan 9th 2013 at 08:25 AM.
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  3. #3
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    Re: General trig question

    Hello, kinhew93!

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

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

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

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