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Posted by on Friday, May 21, 2010 at 1:26pm.

Find the area between y=cosx and y=sinx from 0 to 2pi.

To find the zeros, I combined the equations cosx-sinx=0

What's next?

  • Calculus - , Friday, May 21, 2010 at 4:57pm

    Watch out.
    Between 0 and 45 degrees, the cos is bigger than the sin.
    Between 45 and 90, the sin is bigger than the cos.
    In both sectors, there is area between them. The same in fact.
    Therefore for quadrant 1, do 0 to pi/4 and double the result.
    Then look at the other three quadrants.

  • Calculus - , Friday, May 21, 2010 at 5:05pm

    integral cos x dx from 0 to pi/4 = sin pi/4 - sin 0
    = sqrt(2)/2
    integral sin x dx from 0 to pi/4 = -cos pi/4 + 0 = -sqrt(2) /2
    difference = sqrt 2
    then integral from 0 to pi/2 = 2 sqrt 2
    etc

  • Calculus - , Friday, May 21, 2010 at 5:07pm

    Oh, forgot you do not know where sin = cos

    sure cos x - sin x = 0
    1 -tan x = 0
    tan x = 1
    x = pi/4 (x = y at 45 deg)

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