Known function (FX) = (SiNx cosx) sin2x / SiNx find monotone increasing interval of function FX Why 2K π - π / 2 ≤ 2x - π / 4

Known function (FX) = (SiNx cosx) sin2x / SiNx find monotone increasing interval of function FX Why 2K π - π / 2 ≤ 2x - π / 4

This is using the period of the function y = Sint, which is 2 π
Independent of known functions

Find the maximum value of the function y = - Cos2 square x + SiNx + 2 and the set of X when the maximum value is obtained

y=-(cosx)^2+sinx+2=(sinx+1/2)^2+3/4
When x = - 2K π - π / 6 or x = 2K π - 5 π / 6, the minimum value of Y is 3 / 4
When x = 2K π + π / 2, the maximum value of Y is 3

Find the maximum value of the function y = - Sin square x + 4sinx + 7 divided by 4

Let t = SiNx
Then y = - T & # 178; + 4T + 7 / 4 = - (T-2) &# 178; + 23 / 4
Because | t|

Cos half x times cos half x = half cosx plus half cos2x

cosx/2*cos3x/2
=1 / 2 * [cos (x / 2 + 3x / 2) + cos (x / 2-3x / 2)] (integral sum difference)
=1/2*【cos2x+cos（-x）】
=1/2*（cos2x+cosx）
=1/2*cosx+1/2*cos2x.

If f (x) satisfies f (cosx) = x / 2 (0 ≤ x ≤ π), then f (COS (4 π / 3)) = if f (x) satisfies f (cosx) = cos2x, then the value of F (sin15 °) is

f(cos(4π/3))= f(cos(2π/3))= 2π/3/2=π/3
F (sin15 °) = f (cos75 °) = cos (2 * 75 °) = - (radical 3) / 2

By translating all the points on the image of the function y = cos (2x Pie / 3) to () units, we get the image of the function y = cos 2x, whose period is ()

Left π / 6 π

If f (x) = 3-cosx, then f (SiNx) + F (cos2x)=___
Hope to give detailed steps

f(x)=3-cosx
f(sinx)=3-cos(sinx)
f(cos2x)=3-cos(cos2x)
therefore
f(sinx)+f(cos2x)
=3-cos(sinx)+3-cos(cos2x)
6-cos(sinx)-cos(cos2x)

Given SiNx + siny = 1 / 3, find the maximum and minimum of M = SiNx cosy ^ 2,!

sinx=1/3-siny
So SiNx - (cosy) ^ 2 = (1 / 3-siny) - [1 - (siny) ^ 2]
=(siny)^2-siny-2/3=(siny-1/2)^2-1/4-2/3
siny∈[-1,1],
When siny = - 1, (siny-1 / 2) ^ 2 has a maximum value of 9 / 4,
SiNx - (cosy) ^ 2 gets the maximum value of 4 / 3
When siny = 1 / 2, the minimum value of (siny-1 / 2) ^ 2 is 0,
SiNx - (cosy) ^ 2 gets the minimum value - 11 / 12

Given SiNx + siny = 1 / 2, find the maximum and minimum of cosy's square + 2sinx?

cos^2(y)+2sinx=1-siny*siny+2sinx
=1-(1/2-sinx)*(1/2-sinx)+2sinx
=3/4-(sinx-3/2)^2+9/4
=3-(sinx-3/2)^2
By - 1

Given the function y = 2Sin & # 178; 2x-1, then the function is
Even functions with a period π
Even functions with B period π / 2
Odd functions with C period π
Odd functions with D period π / 2

A:
y=2sin²2x-1=-cos4x
Even function, period T = 2 π / 4 = π / 2
Option B