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Math Help - 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 \cos (x) = 0 then \tan(x)=\infty, and the product of  0 \times \infty is undefined. So no: if  \cos(x)=0, then  \cos(x)[ 1 + \tan(x)] \ne 0.
    Last edited by ebaines; January 9th 2013 at 09:25 AM.
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  3. #3
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    Re: General trig question

    Hello, kinhew93!

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

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

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

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

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

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


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