The known function f (x) = 2sinxcosx + 2cos ^ 2x-1 (1) finding the minimum positive period of function f (x) (2) finding monotone decreasing interval of function f (x) Wait online!

The known function f (x) = 2sinxcosx + 2cos ^ 2x-1 (1) finding the minimum positive period of function f (x) (2) finding monotone decreasing interval of function f (x) Wait online!

f(x)=sin2x+cos2x
=(2 open quadratic) xsin (2x + 45)
So the minimum positive period is 360 / 2 = 180 degrees
The monotone decreasing interval is 90

Let x belong to (0, π / 2), and find the minimum value of the function y = (2Sin & # 178; X + 1) / (sin2x)

1=sin^2x+cos^2x
y=(2sin^2x+sin^2x+cos^2x)/(2sinxcosx)=(3sin^2x+cos^2x)/(2sinxcosx)
y=3/2*tanx+1/2*cotx
Let TaNx = t, Cotx = 1 / T, TaNx = t belong to (0, positive infinity)
Y = (3T + 1 / T) / 2 obviously 3T + 1 / T is the check function 3T + 1 / T > = 2 radical 3
So the minimum value of the function is (2 radical 3) / 2 = radical 3

Let x belong to (0, / 2), then the function y = (2Sin ^ 2 x + 1) / sin2x minimum? To use the discriminant method has been proved to be possible, I will

LZX belongs to (0, / 2) is pi / 2, y = (2Sin ^ 2x + 1) / sin2x = (3tan ^ 2x + 1) / 2tanx let t = TGX belong to (0, positive infinity) u = (3T ^ 2 + 1) / 2T change to quadratic form (quadratic function type, line pronoun) 3T ^ 2-2tu + 1 = 0 as a function of T. u is a parameter variable on (0, positive infinity)

The minimum value of function y = 2Sin (7 / 2 π - 2x) - sin2x?

First, sin (7 / 2 π - 2x) is transformed into - cos2x by using the induced formula "odd variable, even constant, sign quadrant", and then the auxiliary angle formula is used to transform the function into sinusoidal function solution: y = 2Sin (7 / 2 π - 2x) - sin2x = - 2cos2x-sin2x = - √ 5sin (2x + φ) y minimum = - √ 5 number sin^_^
Remember to adopt it

Rescue: (0, PAI), the minimum value of function y = 2Sin ^ 2x + 1 / sin2x?

Did you learn how to find extremum by derivative? Maybe it can be used. I forgot

Let x ∈ (0, π 2), then the minimum value of the function y = 2sin2x + 1sin2x is______ ．

∵ y = 2sin2x + 1sin2x = 2 − cos2xsin2x = k, take the left semicircle of a (0, 2), B (- sin2x, cos2x) ∈ x2 + y2 = 1, as shown in the figure, Kmin = tan60 ° = 3

Given the vector a = (Sina, Sina), B = (1,2) and a * b = 0, ① find the value of Tana, ② find the range of function f (x) = cos & # 178; X + tanasinx, (x ∈ R)
Given the vector a = (Sina, Sina), B = (1,2) and a * b = 0, ① find the value of Tana, ② find the range of function f (x) = cos & # 178; X + tanasinx, (x ∈ R)

(1)sinA+2cosA=0==>sinA= - 2cosA==>tanA= - 2
(2)f(x)=cos²x-2sinx=1-sin²x-2sinx
= －sin²x－2sinx+1= －(sinx+1)²+2
f(max)=2(sinx=-1)
f(min)=- 2 (sinx=1)
The range of F (x) is [- 2,2]

Given the vector M = (Sina, COSA), n = (1, - 2), and m * n = 0.1. Find the value of Tana; 2. Find the range of function f (x) = cos2x + tanasinx (x belongs to R)

(1) So Tana = 0 (2) f (x) = cos2x + tanasinx (2) f (x) = cos2x + tanasinx = (1-2sin2x) + 2sinx = - 2sinx ^ 2x + 2sinx + 1 = - 2 (sinx-1 / 2) ^ 2 + 3 / 2 by SiNx ∈ [- 1,1], then when SiNx = 1 / 2, y max = 3 / 2, SiNx = - 1, y max = - 3, then the range is [- 3,3 / 2]

Given the function f (x) = (SiNx) ^ 4 + (cosx) ^ 4, find the minimum positive period and monotone increasing interval of F (x)

(SiNx) ^ 4 + (cosx) ^ 4 = [(SiNx) ^ 2 + (cosx) ^ 2] ^ 2-2 (SiNx) ^ 2 * (cosx) ^ 2 = 1 ^ 2-2 (SiNx) ^ 2 * (cosx) ^ 2 = 1-2 (sinxcosx) ^ 22sinxcosx = sin2x, so (SiNx) ^ 4 + (cosx) ^ 4 = 1 - (sin2x) ^ 2 / 2 = 1-1 / 2 * 1 / 2 * (1-cos4x) = 1-1 / 4 + 1 / 4cos4x = 3 / 4 + 1 / 4cos4x

Finding the maximum value of the function y = 2 + cosx 1 / 2 + SiNx

Y = (1 + SiNx) / (2 + cosx) find the extreme point: y '= [cosx (2 + cosx) + SiNx (1 + SiNx)] / (2 + cosx) ^ 2 = (2cosx + SiNx + 1) / (2 + cosx) ^ 2 = 0, get: 2cosx + SiNx + 1 = 0 √ 5sin (x + T) = - 1, t = arctan2sin (x + T) = - 1 / √ 5 = - cost = sin (T - π / 2) x + T = t - π / 2 + 2K, or x + T = (2k -