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Sonic
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Vilnius, Lithuania
Dec 2001 time: 07:24
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kos(2x)/(kosx+sinx)dx=(kosx+sinx/2)dx=kosx*dx+sinx/2*dx
Not sure if that is correct, I am not so good at maths...
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Zero-Tau
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Elsewhere
Aug 2002 time: 06:24
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You need the cosine addition formula. It is:
cos(x + y) = cos (x) * cos (y) - sin (x) * sin (y)
Setting x = y, we get:
cos(2x) = (cos(x))^2 - (sin(x))^2
Then we rewrite the expression:
cos(2x) / (cosx + sinx) = ((cos(x))^2 - (sin(x))^2) / (cos(x) + sin(x)) = cos(x) - sin(x)
(For the last step I used the formula (a^2 - b^2) = (a + b)(a - b).)
Now you can evaluate the integral:
(cos(x) - sin(x)) dx = cos(x) dx - sin (x) dx = sin(x) + cos(x)
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spartak
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Why in god's name are you doing trig on Xmas eve?
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