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In the solution to the problem following session 8, it reads "We know that -1 is less than or equal to sin(x) which is less than or equal to 1." I'm good with this, but then it reads: "So it must be true that:" -1/x is less than or equal to sin(x)/x which is less than or equal to 1/x. I don't understand (algebraically) this could be true of all values of x. Can someone please explain?

OCW Scholar - Single Variable Calculus
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If you begin with the inequality stated after "We know that," and multiply every term by 1/x, we get the inequality stated after "So it must be true that." We're just multiplying through by 1/x. You might wonder how we can get away with this, for two reasons. One is the problem at x=0, and the solution acknowledges that we don't have a clear understanding of what happens there. The other is what happens when x turns negative, because multiplying an inequality by a negative number changes the direction of the inequality. That doesn't happen here, which I guess can be explained by the fact that sin(-x)=-sin(x).
Ok. So I understand the algebra now. Thanks! However, I'm a little confused about the "x turns negative" situation. Can you possibly explain this further? Thanks again, TA
I probably caused unnecessary confusion with that remark, just thinking out loud about the general rule that when you multiply an inequality by a negative number, the inequality reverses direction. For example, x>4 implies that -x<-4. That made me wonder why we can multiply by 1/x through this inequality and have it work even when 1/x is negative. It doesn't cause a problem here so we don't have to worry about it.

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Other answers:

Stab in the dark here... could it be that any negative angle can be expressed as a positive angle? I seem to remember this "excuse" being used somewhere else in trig.
I think interchanging negative and positive angle relationships might really depend on the application at hand.

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