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.
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?
3 answers
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
= 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
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)
sure cos x - sin x = 0
1 -tan x = 0
tan x = 1
x = pi/4 (x = y at 45 deg)
1 answer