Given the function f (x) = cosx cos (x + π / 2), X ∈ R, find the maximum of F (x). If f (a) = 3 / 4, find the value of sin2a

Given the function f (x) = cosx cos (x + π / 2), X ∈ R, find the maximum of F (x). If f (a) = 3 / 4, find the value of sin2a

f(x)=sinx+cosx
=√2(√2/2*sinx+√2/2cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)
So the maximum value = √ 2
f(a)=sina+cosa=3/4
square
sin²a+cos²a+2sinacosx=9/16
1+sin2a=9/16
sin2a=-7/16
Given the function f (x) = sin2x-2 (cosx) ^ 2 + 3, find the maximum value of the function and the set of X values when obtaining the maximum value; monotone increasing interval of the function; the set of X satisfying f (x) > 3
f（x）=sin2x-2(cosx)^2+3=sin2x-2(cosx)^2-1+4=sin2x-cos2x+4=√2sin(2x-π/4)+4
So when 2x - π / 4 = π / 2 + 2K π, the maximum value √ 2 + 4 is obtained
Here x = k π + 3 π / 8 (k is an integer)
When - π / 2 + 2K π
f（x）=sin2x-2cos²x+3
=sin2x-(1+cos2x)+3
=sin2x-cos2x+2
=√2(√2/2sin2x-√2/2cos2x)+2
=√2sin(2x-π/4)+2
When 2x - π / 4 = 2K π + π / 2, the maximum value of F (x) is 2 + √ 2
This... Unfolds
f（x）=sin2x-2cos²x+3
=sin2x-(1+cos2x)+3
=sin2x-cos2x+2
=√2(√2/2sin2x-√2/2cos2x)+2
=√2sin(2x-π/4)+2
When 2x - π / 4 = 2K π + π / 2, the maximum value of F (x) is 2 + √ 2
The set of X is {x | x = k π + 3 π / 8, K ∈ Z}
When 2x - π / 4 = 2K π - π / 2, the minimum value of F (x) is 2 - √ 2
The set of X is {x | x = k π - π / 8, K ∈ Z}
From 2K π - π / 2 ≤ 2x - π / 4 ≤ 2K π + π / 2
We obtain K π - π / 8 ≤ x ≤ K π + 3 π / 8, K ∈ Z
The monotone increasing interval of a function is
[kπ-π/8,kπ+3π/8],k∈Z
F (x) > 3 means √ 2Sin (2x - π / 4) + 2 > 3
sin(2x-π/4)>√2/2
∴2kπ+π/4≤2x-π/4≤2kπ+3π/4
∴kπ+π/4≤x≤kπ+π/2,k∈Z
The set of X satisfying f (x) > 3 is
{x | K π + π / 4 ≤ x ≤ K π + π / 2, K ∈ Z} retract
Monotone interval method of logarithmic function
What is the solution of monotone interval of logarithmic function?
What is the decreasing interval of y = Log1 / 2 (- x ^ 2-2x + 3)?
y＝log1/2（-x^2-2x+3)
Take it as a compound function, and the domain is - 3
X is greater than - 3 and less than - 1
Given the function f (x) = sin (2x - π 6) + 2cos2x-1. (I) find the monotone increasing interval of function f (x); (II) in △ ABC, a, B, C are the opposite sides of angles a, B, C respectively, and a = 1, B + C = 2, f (a) = 12, find the area of △ ABC
(II) because f (a) because f (a) because f (a) because f (a) is 12, so the sin (2a + π 6) = 12, so the sin (2a + π 6) = 12 is also 0 < a < a < a < π, so the π 6 is < 2A + π 6 = 12, so the π 6 < 2A + π 6) = 12 is also 0 < a < a < a < π, so the π 6 is < 2A + π 6 < 13 π 6 < 13 π 6 < 13 π 6, so the π 6 is < 2A + 2A + π 6 < 13 π 6 < 13 π 6 < 13 π 6, and thus the π 6 is the corresponding π 6 = 5 π 6 = 5 π 6, so the a = π 6 is a = π 3 in the ABC of the ABC, a = ABC, B = a = ABC, B = a = 1, B = a = 1, B = 1, B + B + C = 1, B + c= π 3  1 = B2 + c2-2bccos A. That is, 1 = 4-3bc, so BC = 1, so s △ ABC = 12bcsina = 34
Is the definition field of logarithmic function the range of X
For example: y = ln x, domain: 0 < x < + ∞ y = ln (x + 2), domain: - 2 < x < + ∞ y = ln (x ^ 2 + 1), domain: - ∞ < x < + ∞ y = LG
What is the original function of (COS x) ^ 4
∫((cosx)^4)=[(cosx)^2]^4=[(1+cos2x)/2]^2
∫(=1/4+(1/2)cos2x+(1/4)(cos2x)^2
∫(=3/8+(1/2)cos2x+(1/8)cos4x
=3x/8+(1/4)sin2x+(1/32)cos4x+C
=3x/8+(1/4)sin2x+(1/32)cos4x+C
Integral (COS x) ^ 4
（cos x）^4=[（cos x）^2]^2=[(1+cos2x)/2]^2
=[1+2cos2x+(cos2x)^2]/4
=1/4+cos2x/2+(1+cos4x)/8
=3/8+cos2x/2+cos4x/8
therefore
The original function is (3 / 8) x + sin2x / 4 + sin4x / 32 + C
Monotone interval of logarithmic function
Monotone interval of y = log4 (x ^ 2 + 2x + 3)
Please write down the process
This is a composite function, so the logarithmic function itself is monotone increasing function, so the monotonicity of the entire composite function depends on the monotonicity of y = x ^ 2 + 2x + 3, then first look at the domain x ^ 2 + 2x + 3 > 0, you can see that the constant holds (the discriminant is less than 0), then look at the monotone interval of y = x ^ 2 + 2x + 3, its monotonicity
Is f (x) = LG 1 = o an odd or even function
Both odd and even functions
f(-x)=f(x)=-f(x)
So it's both odd and even
Monotone interval problem of logarithmic function
How to find monotone interval of logarithmic function?
Find the domain, range and monotone interval of the following functions
1.log1/2 X^2
2.y= loga (-x)
3.y=log2 (X^2-3X-10)
4.y=log1/2 (2-X-X^2)
1. The definition field x is not equal to 0, and the monotone interval from negative infinity to positive infinity is increasing from negative infinity to 0 and decreasing from 0 to positive infinity
2. The domain x is less than 0, the range is from negative infinity to positive infinity, and the monotone interval is 0
X ^ what is this?
Is the function f (x) = LG (A-X / A + x) odd or even?
A>2
(a-x)/(a+x)>0
(a-x)(a+x)>0
(x-a)(x+a)2
-a
Domain (- A, a)
f(-x)=lg(a+x/a-x)
=-lg(a-x/a+x）
=f(x)
So it's an odd function
Odd function - f (- x) = f (x)