Let TaNx = 3, sin ^ 2 x + 2sinxcosx =? From TaNx = 3, SiNx = 3cosx,

Sin? X + 2sinxcosx = (sin? X + 2sinxcosx) / (sin? X + cos? X) divided by cos? X = (tan? 2x + 2tanx) / (tan? 2x + 1) = (3? + 2 * 3) / (3? + 1) = 15 / 10 = 3 / 2

The maximum value of the function y = sin (x + 10 °) + cos (x + 40 °), (x ∈ R) is______ .

The function y = sin (x + 10 °) + cos (x + 40 °)
=sin（x+10°）+cos（x+10°+30°）
=sin（x+10°）+cos（x+10°）cos30°-sin（x+10°）sin30°
=1
2sin（x+10°）+
Three
2cos（x+10°）
=sin（x+70°）
The maximum value of ∵ y = sin (x + 70 °) is 1
The maximum value of the function y = sin (x + 10 °) + cos (x + 40 °) (x ∈ R) is 1
So the answer is: 1

The maximum value of the function y = sin (x + 10 °) + cos (x + 40 °), (x ∈ R) is______ .

Function y = 1 The maximum value of 2 + SiNx + cosx is () A. Two 2−1 B. Two 2+1 C. 1− Two Two D. −1− Two Two

Function y = 1
2+sinx+cosx=1
2+
2sin(x+π
4)
The maximum value is
Two
2+1
Therefore, B is selected

When x ∈ (0, π / 4), the minimum value of 1 / (TaNx Tan ^ 2x)

When x ∈ (0, π / 4), 0

What is the maximum value of (Tan ^ 2x TaNx + 1) / (Tan ^ 2x + TaNx + 1) when x is equal to?

Let a = TaNx, a belong to R
y=(a^2-a+1)/(a^2+a+1)
ya^2+ya+y=a^2-a+1
(y-1)a^2+(y+1)a+(y-1)=0
The discriminant is greater than or equal to 0
(y+1)^2-4(y-1)^2>=0
(y+1+2y-2)(y+1-2y+2)>=0
(3y-1)(-y+3)>=0
(3y-1)(y-3)

If TaNx = 2, then Tan (π / 4 + 2x) =?

First, we calculate tan2x = 2tanx / 1-tanx ^ 2 = 2 * 2 / 1-2 ^ 2 = - 4 / 3
tan(π/4+2x)=(tan2x+tanπ/4)/1-(tanπ/4*tan2x)=(-4/3+1)/1-(-4/3)=-1/7

The function y = tanx-tan3x The product of the maximum and minimum of 1 + 2tan2x + tan4x is___ .

∵y=tanx-tan3x
1+2tan2x+tan4x=tanx(1-tan2x)
(1+tan2x)2=tanx
1+tan2x•1-tan2x
1+tan2x
=1
2sin2x•cos2x=1
4sin4x，
Therefore, the maximum and minimum values are: 1
4 and - 1
Four
The product of maximum and minimum is - 1
Sixteen
The answer is as follows:
Sixteen

Y = 2tan ^ 2x / (1-tanx), X belongs to (PAI / 4, Pai / 2), and the maximum value of the function is?

A > 1y = 2A ^ 2 / (1-A) = [2 (a ^ 2-1) + 2] / (1-A) = 2 (a ^ 2-1) + 2] / (1-A) = 2 (a ^ 2-1) / (1-A) + 2 / (1-A) = 2 (a ^ 2-1) / (1-A) + 2 / (1-A) = - 2 (a + 1) + 2 * (1-A) = - 2 (A-1) - 4 + 2 / (1-A) = - 2 [(A-1) + 1 / (A-1)] - 4A > 1, A-1 > 0 (A-1) + 1 / (A-1) > = 2 root number (A-1) (A-1) (A-1) / (A-1) > = 2 root number (A-1) / (A-1) - 1) = 2, so - 2 [(a - 1) + 1 / (a - 1

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The purpose of the question should be y = tan2x * TaNx * TaNx * TaNx
That's y = tan2x (TaNx) ^ 3
y=tan2x(tanx)^3
=2tanx/[1-(tanx)^2]*(tanx)^2
=2(tanx)^4/[1-(tanx)^2]
=-2[(tanx)^4-(tanx)^2+(tanx)^2-1+1]/[(tanx)^2-1]
=-2{(tanx)^2+1+1/[(tanx)^2-1]}
=-2{[(tanx)^2-1]+1/[(tanx)^2-1]+2}
Because pi / 41 --- > (TaNx) ^ 2 > 1 --- > (TaNx) ^ 2-1 > 0
According to the mean inequality [(TaNx) ^ 2-1] + 1 / [(TaNx) ^ 2-1] > = 2
--->[(tanx)^2-1]+1/[(tanx)^2-1]+2>=4
--->-2{[(tanx)^2=1]+1/[(tanx)^2-1]+2}=<-8
And (TaNx) - 1 = 1 / [(TaNx) ^ 2-1] < --- > [(TaNx) ^ 2-1] = 1 --- > (TaNx) ^ 2 = 2
pi/4tanx=√2
So when x = arctan √ 2, y = tan2x (TaNx) ^ 3 has a maximum of - 8

Let TaNx = 3, sin ^ 2 x + 2sinxcosx =? From TaNx = 3, SiNx = 3cosx,