Find the function f (x) = cos ^ 2x SiNx, X belongs to the maximum value of [- π / 4, π / 4]

Find the function f (x) = cos ^ 2x SiNx, X belongs to the maximum value of [- π / 4, π / 4]

f(x)=cos²x-sinx
=1-sin²x-sinx
=-(sin²x+sinx)+1
=-(sinx+1/2)²+5/4≤5/4
∵x∈[-π/4,π/4]
∴sinx∈[-√2/2,√2/2]
/ / SiNx can be taken as - 1 / 2, that is, the equal sign can be taken
So the maximum is 5 / 4
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What is the maximum value of the function y = 2sinx (SiNx + COS)?

Original = 2 (SiNx) ^ 2 + 2sinxcosx
Because 2 (cosx) ^ 2-1 = cos2x
2 (cosx) ^ 2 = 2-2 (SiNx) ^ 2; 2 (SiNx) ^ 2 = 2-2 (cosx) ^ 2
Y = 2-2 (cosx) ^ 2 + sin2x = 1-2 (cosx) ^ 2 + 1 + sin2x = 1-cos2x + sin2x = 1 + radical 2 × sin (2x-45)
The maximum value is 1 + root 2

Find the maximum and minimum values of the function y = cos? X + 2sinx-2

The solution is determined by y = cos ^ 2x + 2sinx-2
=1-sin^2x+2sinx-2
=-sin^2x+2sinx-1
=-（sinx-1）^2
So when SiNx = 1, y has a maximum value of 0
When SiNx = - 1, y has a minimum value = - 4

Find the maximum and minimum values of the function y = SiN x + cos x + 2sinx cos x + 4 I still don't understand this question after I read the answer Let k = SiN x + cos x, then SiNx cosx = (k ^ 2 - 1) / 2 -------- / / I know what's going on in this step Because: SiN x + cos x = √ 2 sin (x + pi / 4) So: K ∈ [- √ 2, √ 2] Y = K + K ^ 2 - 1 + 4 = k ^ 2 + K + 3 = (K + 1 / 2) ^ 2 + 11 / 4 -------- / / I understand this step When k = - 1 / 2, y min = 11 / 4; when k = √ 2, y max = 5 + √ 2 / / I understand this step too I don't understand the two steps in the middle

Two steps in the middle:
Because: SiN x + cos x = √ 2 sin (x + π / 4)
So: K ∈ [- √ 2, √ 2]
[analysis]
sinx＋cosx
＝√2(sinx•√2/2＋cosx•√2/2)
＝√2(sinx•cosπ/4＋cosx•sinπ/4)
＝√2sin(x＋π/4)
∵－1≤sin(x＋π/4)≤1
∴－√2≤√2sin(x＋π/4)≤√2
Thus: K ∈ [- √ 2, √ 2]
The function y = asinx + bcosx can be transformed as follows:
y＝√(a²＋b²) sin(ωx＋φ)．
last,

Let f (x) = cos ^ 2x-2sinx obtain the minimum and maximum values of the independent variable x

f(x)=cos^2x-2sinx
=1-(sinx)^2-2sinx
=-(sinx+1)^2+2
When SiNx = - 1, i.e. x = 3 π / 2 + 2K π, the maximum value is 2
When SiNx = 1, i.e. x = π / 2 + 2K π, the minimum value is - 2

The range of y = SiNx + cos? X

y=sinx+1-sin²x
=-（sinx-1/4）²+17/16
SiNx range: [- 1,1]
The maximum value is: 17 / 16
Minimum: - 1

Y = cos? X-sinx + 1 x belongs to the range of π / 3 to 3 π / 4

y=1-sin²x-sinx+1=2-(sinx+1/2)²+1/4=9/4-(sinx+1/2)²
X ∈ [π / 3,3 π / 4] so SiNx ∈ [√ 2 / 2,1] so the range [0,3 / 2 - √ 2 / 2]

If | x|

(x) = cos ^ 2x + SiNx = 1-sin ^ 2x + SiNx = 1-sin ^ 2x + SiNx = - (sinx-1 / 2) ^ 2 + 5 / 4

What is the minimum value of the function f (x) = (cosx) ^ 2 + SiNx on the interval [- π / 4, π / 4]

f(x)=cos²x+sinx
=1-sin²x+sinx
=－(sinx-1/2)²+5/4
∵x∈[-π/4,π/4]
∴sinx∈[－√2/2,√2/2]
The axis of symmetry is SiNx = 1 / 2
When sin = - 2 / X is far away from the π - axis
The minimum value of F (x) is obtained: cos 2 (- π / 4) + sin (- π / 4) = 1 / 2 - √ 2 / 2
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Given the vector a = (cos3x / 2, - sin3x / 2), B = (cosx / 2, SiNx / 2), X ∈ [0, π / 2], if the function f (x) = a · B-1 / 2, the minimum value of absolute value a + B is - 3 / 2, find the value of real number λ If the value of λ is 2, the process should be complete.

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