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    Im going through C3 and something twisted my brain. We are told that you can't divide by sin x or cos x in the interval 0 to 360 because it may be 0 making the equation undefined however in many cases when we have sinx=cos x we solve for tan x = 1 in that interval and when i check the solution bank they sometimes multiply both sides by cos x even tho the interval is 0 to 180.
    The question here is when can I know whether or not to divide by cos x or sin x?
    Because i havent been able to find a solid rule for that
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    (Original post by Tirawi19)
    Im going through C3 and something twisted my brain. We are told that you can't divide by sin x or cos x in the interval 0 to 360 because it may be 0 making the equation undefined however in many cases when we have sinx=cos x we solve for tan x = 1 in that interval and when i check the solution bank they sometimes multiply both sides by cos x even tho the interval is 0 to 180.
    The question here is when can I know whether or not to divide by cos x or sin x?
    Because i havent been able to find a solid rule for that
    I'll be working with the range 0 \leq x \leq 2\pi for this answer. If you want to be really careful, when you have the equation \sin x=\cos x, you can separate it into two cases: when \cos x=0 and when \cos x\neq0. If \cos x = 0, then x=\frac{\pi}{2}, \frac{3\pi}{2}, but \sin \frac{\pi}{2} \neq 0 and \sin \frac{3\pi}{2} \neq 0 so there are no solutions when \cos x=0. Now consider the case when \cos x \neq 0 . When this is the case, you can divide the equation by \cos x and proceed as normal. I think it's great that you're thinking carefully about what you're doing, and whether it's valid or not.
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    (Original post by Tirawi19)
    Im going through C3 and something twisted my brain. We are told that you can't divide by sin x or cos x in the interval 0 to 360 because it may be 0 making the equation undefined however in many cases when we have sinx=cos x we solve for tan x = 1 in that interval and when i check the solution bank they sometimes multiply both sides by cos x even tho the interval is 0 to 180.
    The question here is when can I know whether or not to divide by cos x or sin x?
    Because i havent been able to find a solid rule for that
    Good question!


    If a trig function could be equal to 0 in an equation then you should not divide by it e.g.

    \sin x \cos x = \cos x

    If you divide by \cos x you get

    \sin x = 1

    But if you factorise instead you get

    \cos x \left(\sin x - 1\right) = 0

    Which leads to \cos x = 0 as well as \sin x = 1. So dividing by \cos x loses the solution \cos x = 0.

    If you're dividing an equation by cos to give tan, for the majority of occasions you'll be fine since \cos x can't be equal to 0 in your equation. But it's always worth checking.

    So for \cos x = \sin x, it is not possible for \cos x to be 0 and for that to satisfy the eqaution (think about this and let us know if you need more explanation). So it's fine to divide the equation by \cos x.
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    (Original post by I hate maths)
    I'll be working with the range 0 \leq x \leq 2\pi for this answer. If you want to be really careful, when you have the equation \sin x=\cos x, you can separate it into two cases: when \cos x=0 and when \cos x\neq0. If \cos x = 0, then x=\frac{\pi}{2}, \frac{3\pi}{2}, but \sin \frac{\pi}{2} \neq 0 and \sin \frac{3\pi}{2} \neq 0 so there are no solutions when \cos x=0. Now consider the case when \cos x \neq 0 . When this is the case, you can divide the equation by \cos x and proceed as normal. I think it's great that you're thinking carefully about what you're doing, and whether it's valid or not.
    Thanks alot u really helped me get through so much !
 
 
 
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