When 0 < x < π 4, the minimum value of the function f (x) = cos2xcosxsinx − sin2x is______ ．

When 0 < x < π 4, the minimum value of the function f (x) = cos2xcosxsinx − sin2x is______ ．

If f (x) = 1tanx − tan2x ∵ 0 ＜ x ＜ π 4, 0 ＜ TaNx ＜ 1, TaNx − tan2x = − (TaNx − 12) 2 + 14, TaNx = 12, the maximum value of tanx-tan2x is 14, so the minimum value of F (x) = cos2xcosxsinx − sin2x is 4, so the answer is 4

Given that 0 ＜ LG [11 / 2-9cos (x + Pie / 6)] ≤ 1, find the minimum value of the function y = Tan & # 178; x-2tanx + 5

From 0 ＜ LG [11 / 2-9cos (x + π / 6)] ≤ 1
Conclusion: 1

What is the minimum value of 2tanx + Tan (π / 2-x)! X is between (0 - π)
How to calculate!

Original formula = 2tanx + Cotx = 2tanx + 1 / TaNx
0