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    Why does (sin 2x) / (cos 2x) = tan 2x . I thought the answer was tan x because the 2s cancel each other out
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    (Original post by man111111)
    Why does (sin 2x) / (cos 2x) = tan 2x . I thought the answer was tan x because the 2s cancel each other out
    nah, g math don't work like that.

     \sin () is a function.

     \cos() is a function

    You are applying  2x to that function, you can't just cancel the (2x). Imagine the trig functions as vacuum cleaners, once you suck something up with the vacuum cleaner you can't get it back in it's purest form.
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    if it was 2sin x and 2cos x you could've cancel the 2s out, ah man i do miss a level maths :')
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    As a more general formula

     \frac{\sin()}{\cos()} = \tan()

    So....

     \frac{\sin(1)}{\cos(1)} = \tan(1)

     \frac{\sin(x)}{\cos(x)} = \tan(x)

     \frac{\sin(2x)}{\cos(2x)} = \tan(2x)

     \frac{\sin(3x)}{\cos(3x)}= \tan(3x)

    In words, Sine of anything, divided by cosine of that same anything equals tangent of that same anything

    Have I left anything out?

    nah that's moist
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    (Original post by Apachai Hopachai)
    As a more general formula

     \frac{\sin()}{\cos()} = \tan()

    So....

     \frac{\sin(1)}{\cos(1)} = \tan(1)

     \frac{\sin(x)}{\cos(x)} = \tan(x)

     \frac{\sin(2x)}{\cos(2x)} = \tan(2x)

     \frac{\sin(3x)}{\cos(3x)}= \tan(3x)

    In words, Sine of anything, divided by cosine of that same anything equals tangent of that same anything

    Have I left anything out?

    nah that's moist
    That "anything" cannot equal \frac{\pi}{2}+\pi n for n\in \mathbb{Z} otherwise you get division by 0
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    (Original post by RDKGames)
    That "anything" cannot equal \frac{\pi}{2}+\pi n for n\in \mathbb{Z} otherwise you get division by 0
    yes
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    (Original post by RDKGames)
    That "anything" cannot equal \frac{\pi}{2}+\pi n for n\in \mathbb{Z} otherwise you get division by 0
    you had to confuse it..
 
 
 
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