Prove that xcosx< sinx for 0<x<(5π/4)
-differentiate the function f(x)=sinx-xcosx
If we let f(x) = sinx - xcosx,
f(0) = 0
f'(x) = cos x - cosx + xsinx = xsinx
Now, f(5π/4) = -1/√2 + 5π/4 * 1/√2
= 1/√2 (5π/4 - 1)
so, f(x) is still > 0 at the end of the interval
f'(x) = 0 at x=π, so that's a maximum
so, since f is decreasing on [π,5π/4], but still positive, I'd say that
xcosx < sinx on the whole interval.
To prove that xcosx < sinx for 0 < x < (5π/4), we need to show that the function f(x) = sinx - xcosx is positive in the given interval.
To do this, we can differentiate the function f(x) and analyze its behavior.
1. Differentiating the function:
f(x) = sinx - xcosx
Applying the product rule:
f'(x) = d/dx(sinx) - d/dx(xcosx)
= cosx - (1)(cosx) - x(-sinx)
= cosx - cosx + xsinx
= xsinx
2. Analyzing the sign of the derivative:
To determine whether f(x) is increasing or decreasing in the given interval, we need to analyze the sign of f'(x).
When xsinx > 0:
Since sinx is positive for 0 < x < (π/2), and x is also positive because 0 < x < (5π/4), the product xsinx will be positive.
Therefore, f(x) is increasing when xsinx > 0.
3. Conclusion:
Since f'(x) = xsinx is positive in the interval 0 < x < (5π/4), we can conclude that f(x) = sinx - xcosx is also increasing in this interval.
Since f(0) is negative (plug in x = 0: f(0) = sin(0) - 0cos(0) = 0), and f(x) is increasing, we can infer that f(x) remains negative for all x in the interval 0 < x < (5π/4).
Hence, we have proved that xcosx < sinx for 0 < x < (5π/4).