Thread: 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!

Hello, Sconts!

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

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·sin[(A+B)/2]·sin[(A-B)/2]


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

. . which becomes: .-2·sin(5x)·sin(-x) .> .0

And we have: .2·sin(5x)·sin(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!

Thank you very much for the help, you saved me

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 )

Once again, thanks
Good method!