# Trig Inequality

• Feb 15th 2007, 08:19 PM
Sconts
Trig Inequality
cos(4x) > cos(6x)

I've tried to let A = 2x, and solve it using duoble angles...

cos(2A) > cos(2A+A) ...then letting cos2AcosA - sin2AsinA replace cos(2A+A), etc. but it gets really messy and I get lost.

Can anyone please do this step by step?

Thanks in advance! :)
• Feb 15th 2007, 09:35 PM
Soroban
Hello, Sconts!

I found an approach, but it's just as bad as your expansion . . . LOL!

Quote:

cos(4x) .> .cos(6x)
I'll assume that the answer are between 0 and 2π.

I used a sum-to-product identity: . cos A - cos B .= .-2新in[(A+B)/2]新in[(A-B)/2]

We have: .cos(4x) - cos(6x) .> .0

. . which becomes: .-2新in(5x)新in(-x) .> .0

And we have: .2新in(5x)新in(x) .> .0

There are two cases to consider:
. . [1] Both factors are positive.
. . [2] Both factors are negative.

[1] sin(x) > 0 .and .sin(5x) > 0

If sin(x) > 0, then: .0 < x < π

If sin(5x) > 0, then:
. . 0 < 5x < π . . . . . .0 < x < π/5
. .2π < 5x < 3π . . 2π/5 < x < 3π/5
. .4π < 5x < 5π . . 4π/5 < x < π

Hence: .x ε (0, π/5) U (2π/5, 3π/5) U (4π/5, π)

[2] sin(x) < 0 .and .sin(5x) < 0

If sin(x) < 0, then: .π < x < 2π

If sin(5x) < 0, then:
. . 5π < 5x < 6π . . . . . .π < x < 6π/5
. . 7π < 5x < 8π . . . 7π/5 < x < 8π/5
. . 9π < 5x < 10π . . 9π/5 < x < 2π

Hence: .x ε (π, 6π/5) U (7π/5, 8π/5) U (9π/5, π)

The solution is the union of all the intervals in blue.

But there must be a better way!
• Feb 16th 2007, 03:13 AM
Sconts
Thank you very much for the help, you saved me :D

I kept using half angle and cos(a+B) formulas and it got wayyy too messy, haha (I probably just didn't know how to simplify it, but okay :p)

Once again, thanks :)
Good method!