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  1. #1
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    Solve

    Solve sin x > cos x . Sucks don't know how to do it.Somebody please help
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  2. #2
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    1st you divide both left and right by cos x which gives you:

    (sin x/cos x) > (cos x/cos x)

    and we recognize that the left term will now be tan x

    tan x > 1
    x > arc tan 1
    x > 45 degree
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  3. #3
    o_O
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    Not true. Take x = \pi = 180^{o} > 45^{o} yet we have that \tan 180^{o} = 0 < 1

    Imagining the unit circle would help and then generalize the case.
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  4. #4
    Moo
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    Hello,

    You must differentiate the case when cos(x) is < 0, in which case the inequality will change its direction.

    And as tan is periodic, let's restrict this to [0;2pi[
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  5. #5
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    given - \pi <x> \pi
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  6. #6
    Member Danshader's Avatar
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    hmm good point lemme see,

    tan x > 1

    whereby tangent x is positive (shown on right hand side by the digit 1) hence we should consider the 1st and 3rd quardrant to find its limits.

    x = 45 degree (lower limit)
    x = 225 degree (upper limit)

    hence the range it can go is

    45 < x < 225 (if we only consider what the question wants and by inputing values into the calculator and comparing for sin and cos)

    if we were to consider tan x>1 then the limits will be:
    45 < x < 90
    and
    180 < x < 225
    Last edited by Danshader; March 30th 2008 at 10:30 AM. Reason: sounds better
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  7. #7
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    Quote Originally Posted by Danshader View Post
    1st you divide both left and right by cos x which gives you:

    (sin x/cos x) > (cos x/cos x)

    and we recognize that the left term will now be tan x

    tan x > 1
    x > arc tan 1
    x > 45 degree
    when you arc tan it, the domain is restricted to π/2>=x>=-π/2
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