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Math Help - Trignometry qus

  1. #1
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    Trignometry qus

    Find all values of x in the interval 0≤X2 pie

    a) sin x = cos x




    Could you please provide me working for it, thanks.
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  2. #2
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    Ignore thread please

    Sorry, dont worry about working the qus out..

    I just figured out how to do it, thanks
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tom_asian View Post
    Find all values of x in the interval 0≤X2 pie

    a) sin x = cos x




    Could you please provide me working for it, thanks.
    well, you should pretty much know by heart that \sin x = \cos x for x = \frac {\pi}{4} + k \pi

    but if you wanted to work it out explicitly, here's what to do

    \sin x = \cos x

    \Rightarrow \sin^2 x = \cos^2 x

    \Rightarrow \cos^2 x - \sin^2 x = 0

    \Rightarrow \cos 2x = 0 ........you should also know for what values \cos x = 0

    \Rightarrow 2x = \frac {\pi}{2} + 2k \pi

    \Rightarrow x = \frac {\pi}{4} + k \pi

    Now i leave the easy part for you. what are the values in the interval we were given?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tom_asian View Post
    Sorry, dont worry about working the qus out..

    I just figured out how to do it, thanks
    couldn't you have said that sooner?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    well, you should pretty much know by heart that \sin x = \cos x for x = \frac {\pi}{4} + k \pi

    but if you wanted to work it out explicitly, here's what to do

    \sin x = \cos x

    \Rightarrow \sin^2 x = \cos^2 x

    \Rightarrow \cos^2 x - \sin^2 x = 0

    \Rightarrow \cos 2x = 0 ........you should also know for what values \cos x = 0

    \Rightarrow 2x = \frac {\pi}{2} + 2k \pi

    \Rightarrow x = \frac {\pi}{4} + k \pi

    Now i leave the easy part for you. what are the values in the interval we were given?
    Ah! But cos(2x) = 0 for
    x = \frac{ \pi}{2} + 2k \pi \text{ and } \frac{3 \pi}{2} + 2k \pi

    So
    x = \frac{ \pi}{4} + k \pi \text{ and } \frac{3 \pi}{4} + k \pi

    It is just that the 3 \pi /4 + k \pi solution is extraneous in this case, its origin coming from squaring both sides of the original equation.

    -Dan
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  6. #6
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    And easier solution.
    \sin x = \cos x
    Divide by \cos x to get,
    \tan x = 1 now continue...
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