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If you want to study the monotony of the function, we'll have to use the derivative of the function. We'll differentiate the function with respect to x:
f'(x) = cos x - sinx
Now, we'll cancel f'(x):
`f'(x) = 0 lt=gt cos x - sinx = 0 =gt 1 - sin x/cos x = 0`
But `sinx/cosx = tan x`
1 - tan x = 0
tan x = 1
The values of the tangent function are positive within the 1st and the 3rd quadrants.
Therefore, we'll have:
`x = pi/4`
`` `x = pi + pi/4`
`` `x = (5pi)/4`
We notice that the tangent function is increasing over the interval (0,pi/2), therefore the function f(x) is increasing over (0,pi/2). Since the values of the tangent function are negative over (pi/2,pi), the derivative f'(x) is decreasing over this interval and the function f(x) is also decreasing over the interval (pi/2;pi).
Therefore, the function is increasing over (0;pi/2) and (pi ; (3pi)/2) and it is decreasing over (pi/2 ; pi) and ((3ipi)/2 ; 2pi).
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