Trigonometry inequation with absolute values

[Johulus]

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \)

So I don't have an idea how to solve this.

I thought maybe I could do something like this

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \\ 2 \sin|x|\, \cdot\, \cos(x)\, >\, \dfrac{1}{2} \\ 4 \sin^2|x|\, \cdot\, \cos^2(x)\, >\, \dfrac{1}{4} \)

I though I could somehow get from that:

\(\displaystyle \sin^2(2x)\, >\, \dfrac{1}{4} \)

But, I don't know what about absolute value. Any tips?

 

[stapel]

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \)

So I don't have an idea how to solve this.

I thought maybe I could do something like this

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \\ 2 \sin|x|\, \cdot\, \cos(x)\, >\, \dfrac{1}{2} \\ 4 \sin^2|x|\, \cdot\, \cos^2(x)\, >\, \dfrac{1}{4} \)

This is fine. Now apply the double-angle identity:

. . . . .\(\displaystyle \sin\left(2x\right)\, >\, \dfrac{1}{2}\)

For what basic reference angle value is the sine equal to one-half?

I though I could somehow get from that:

\(\displaystyle \sin^2(2x)\, >\, \dfrac{1}{4} \)

No, you can't square both sides of an inequality. For instance:

. . . . .\(\displaystyle -2\, <\, 1\)

...but:

. . . . .\(\displaystyle 4\, =\, (-2)^2\, >\, 1^2\, =\, 1\)

But, I don't know what about absolute value.

Try using the properties of the sine function. Isn't is an odd function, so sin(-x) = -sin(x)? And, for x < 0, isn't |x| = -x? So sin(|x|), for x > 0, is just sin(x); but sin(|x|), for x < 0, is sin(-x) = -sin(x). Then proceed in the usual manner.

 

[Steven G]

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \)

So I don't have an idea how to solve this.

I thought maybe I could do something like this

\(\displaystyle \sin|x|\, \cdot\, \cos(x)\, >\, 0.25 \\ 2 \sin|x|\, \cdot\, \cos(x)\, >\, \dfrac{1}{2} \\ 4 \sin^2|x|\, \cdot\, \cos^2(x)\, >\, \dfrac{1}{4} \)

I though I could somehow get from that:

\(\displaystyle \sin^2(2x)\, >\, \dfrac{1}{4} \)

But, I don't know what about absolute value. Any tips?

You need to know that the absolute value function is a piecewise function.

|x| = x if x>=0 and |x| = -x if x<0

For example, |x- 3| = x-3 if x-3>0, ie x>=3 and |x-3| = -(x-3) = 3-x if x-3 < 0, ie if x < 3.

sin|x| = sinx if x>=0 and sin|x| = sin (-x) = -sin x if x<0.

So consider you problem in two cases.

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Case 1: x>=0 AND sinx*cosx > 0.25

Case 2: x<0 AND -sinx*cosx > 0.25 (or sinx*cosx<-0.25)


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