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Homework Help: Trigo doubt

  1. Nov 9, 2005 #1
    My doubt lies with the following question. It is given that sin(pi*cosx) = cos(pi*sinx).
    From the following information it is to be proved that x = (1/4)[(2n+1)pi +or- cos^-1(1/8)] where n is any natural number.



    From the question we can make it clear that the range of pi*cosx and pi*sinx is between pi and -pi. So I took two cases. In the first case bot are positive. When they are positive they must lie in [0,pi/2]. Thus from the equation we get that pi*cosx and pi*sinx must be complementary.
    Therefore pi*cosx = pi/2 - pi*sinx. Therefore cosx = (1/2) - sinx. Therefore sinx + cosx = (1/2) and this is the maxima of the function thus bringing up that x is 2n*pi + pi/4 in case1 while it is 2n*pi -3pi/4 in case2.
    So that is something that is against my question or infact I have disproved what I should prove. Can anyone help?
     
  2. jcsd
  3. Nov 10, 2005 #2
    Just want to clear something up:
    At University, we are seriously discouraged from writing, Sin^-1 /
    Cos^-1 / Tan^-1, because they can mean two things.

    First it, in your question, it could equal arccos(1/8):
    (The angle created from a right-angled triangle with adjacent length 1 & hypotenuse length 8).

    Second it can equal sec(1/8):
    (The inverse of the value cos(1/8) i.e. (Cos 1/8)^-1

    Obviously this is a special case, and you can write Sin^2(x) without confusion. Could you clarify which cos the question is referring to?

    Regards,
    Sam
     
  4. Nov 10, 2005 #3

    VietDao29

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    Homework Helper

    The bolded part is wrong!
    Are you sure that:
    [tex]\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{2}[/tex]??
    In fact:
    [tex]\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} = \sqrt 2[/tex]
    To solve for x in:
    [tex]\sin x + \cos x = \frac{1}{2}[/tex], you can just use the identity:
    [tex]\sin x + \cos x = \sqrt 2 \cos \left( x - \frac{\pi}{4} \right)[/tex].
    So:
    [tex]\sin x + \cos x = \frac{1}{2}[/tex]
    [tex]\Leftrightarrow \sqrt 2 \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2}[/tex]
    [tex]\Leftrightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2 \sqrt 2}[/tex].
    You should be able to go from here, right?
     
  5. Nov 10, 2005 #4
    sinx + cosx = 2^(1/2)cos(x-pi/4)
    is this the modification of any standard identity or something of university level. i cannot recall it from something i learnt.
     
  6. Nov 11, 2005 #5

    VietDao29

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    Homework Helper

    A short proof:
    [tex]\sin x + \cos x = \sqrt 2 \left( \frac{1}{\sqrt 2} \sin x + \frac{1}{\sqrt 2} \cos x \right) = \sqrt 2 \left( \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x \right) = \sqrt 2 \cos \left( x - \frac{\pi}{4} \right)[/tex]
    ---------------
    To solve trig equation in form : a sin x + b cos x = c. Where a, b, c are constants. We often divide boths sides by [itex]\sqrt{a ^ 2 + b ^ 2}[/itex]
    Then let:
    [tex]\sin \alpha = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \quad \mbox{and} \quad \cos \alpha = \frac{b}{\sqrt{a ^ 2 + b ^ 2}}[/tex]
    Or let:
    [tex]\cos \alpha = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \quad \mbox{and} \quad \sin \alpha = \frac{b}{\sqrt{a ^ 2 + b ^ 2}}[/tex]
    So:
    [tex]\sin x + \cos x = \frac{1}{2}[/tex]
    [tex]\Leftrightarrow \frac{1}{\sqrt 2} \sin x + \frac{1}{\sqrt 2} \cos x = \frac{1}{2 \sqrt 2}[/tex]
    [tex]\Leftrightarrow \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x = \frac{1}{2 \sqrt 2}[/tex]
    [tex]\Leftrightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2 \sqrt 2}[/tex]
     
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