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Show that $ \sin x < x $ if $ 0 < x < 2\pi $.

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01:59

Fahad Paryani

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 2

The Mean Value Theorem

Derivatives

Differentiation

Volume

Missouri State University

Campbell University

Harvey Mudd College

Baylor University

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in this problem, we want to show that sine of X is less than X on the interval. Uh Where zero is less than X, which in turn is less than two pi. So X is in this interval from zero, not including zero up to two pi, not including two pi. So on this interval where X is between zero and two pi, we want to show that sine of X is less than X. If sine of X is less than X, that means X is greater than sign of X. And if X is greater than sign of X then x minus sine of X would have to be greater than zero. So we can show that X is greater than sign of X on this interval. Uh huh. We can show that X is greater than sign of X on this interval between zero and two pi. Uh If we can show that X minus sine of X is positive on this interval between zero and two pi remember we're not including zero because at the point X equals zero X and sine of X um are equal to each other. Uh But for X values greater than zero and less than two pi, you can see graphically uh that the function x minus sine of X is a positive function because it's graph stays above the X axis. So X minus sine of X is positive, which means X must be larger or greater than sign effects

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