For \(\displaystyle{0}{ < }{x}{ < }{\frac{{\pi}}{{{4}}}}\) prove

Cazzaoro0w9

Cazzaoro0w9

Answered question

2022-04-14

For 0<x<π4 prove that cosxsin2x(cosxsinx)>8

Answer & Explanation

jchordig1d5

jchordig1d5

Beginner2022-04-15Added 7 answers

Let me try. We have
cosxsin2x(cosxsinx)=1+tan2xtan2x(1tanx)
Note that 0<x<π4, so 0<tanx<1,
tanx(1tanx)14(tanx+1tanx)2=14
So we have
LHS41+tan2xtanx8
The equality happens when tanx=12 for first and tanx=1 for second, so LHS>8.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?