Sin3x + cos3x = (SiNx + cosx) (1-sinxcosx) intermediate simplification process

Sin3x + cos3x = (SiNx + cosx) (1-sinxcosx) intermediate simplification process

sin3x+cos3x=sin(x+2x)+cos(x+2x)=sinxcos2x+cosxsin2x+cosxcos2x-sinxsin2x=sinx(2cos²x-1)+2sinxcos²x+cosx(1-2sin²x)-2sin²xcosx=4sinxcos²x-sinx+cosx-4sin²xcosx=(cosx-sinx)(4sinxcosx+1）
The result is not (SiNx + cosx) (1-sinxcosx)
3x equals x + 2x
Let a and B belong to R, and a is not equal to 2. The function f (x) = LG (1 + ax) / (1 + 2x) defined on the interval (- B, b) is an odd function, then the value range of B
Odd function f (- x) = - f (x) f (x) + F (- x) = 0ln [(1 + ax) / (1 + 2x)] + ln [(1-ax) / (1-2x)] = 0ln {[(1 + ax) / (1 + 2x)] [(1-ax) / (1-2x)]} = 0 = ln1 [(1 + ax) / (1 + 2x)] [(1-ax) / (1-2x)] = 1 (1 + ax) (1-ax) = (1 + 2x) (1-2x) 1-A & sup2; X & sup2; = 1-4x & sup2; so a & sup2
The range of function y = x-4x-1?
Can such a condition be worked out?
y=x²-4x-1
=（x-2）²-5
∵（x-2）²≥0
∴y≥-5
The value range (- 5, + ∞)
The formula is y = x ^ 2-4x + 4-4-1
=(x-2)^2-5
So y is greater than or equal to - 5
Greater than or equal to minus five
Yes, every quadratic equation with one variable has a maximum value. When x = - 2, it has a minimum value, - 5, so the range is - 5 to positive infinity, left closed and right open
Formula y = x ^ 2-4x + 4-4-1 = (X-2) ^ 2-5
Because (X-2) ^ 2 is greater than or equal to 0, then y is greater than or equal to - 5, X belongs to - 5 (can be equal to) to positive infinity
Given that y = f (x) is an odd function, when x is greater than or equal to 0, f (x) = x (1-x), then when x is less than 0, find the analytic expression of F (x)?
Y = f (x) is an odd function. When x ≥ 0, f (x) = x (1 + x),
When X0
∴f(-x)=-x(1-x),
∵ f (x) is an odd function
∴-f(x)=-x(1-x),
f(x)=x(1-x),
When x
In a mathematical problem, the formula method is used to find the maximum value and range of the following functions 1 y = x2 + 4x + 3 x belongs to [- 4, - 3] x belongs to [3,5] x belongs to [- 3,5]
Thank you. Thank you for being more detailed. Thank you
y = (x + 2)^2 - 1
The image is symmetric about x = - 2
The vertex coordinates are (- 2, - 1)
So:
[- 4, - 3]: the value range is: [0,3]
[3,5]: the range is: [24,48]
[- 3,5]: the range is: [- 1,48]
Given that y = (x) is an odd function, when x is greater than or equal to 0, f (x) = x (1-x), find the analytic expression of function f (x)
∵ y = f (x) is an odd function, ∵ f (x) = - f (- x)
If x0, then f (- x) = - x · (1 - (- x)) = - x (x + 1) = - f (x), that is, f (x) = x (x + 1)
∴f(x)=x(1-x) (x≥0)
=x(x+1) (x
∵ y = f (x) is an odd function, ∵ f (x) = - f (- x)
If x0, then f (- x) = - x · (1 - (- x)) = - x (x + 1) = - f (x), that is, f (x) = x (x + 1)
∴f(x)=x(1-x) (x≥0)
=x(x+1) (x
Finding the range of function y = sin (x + 20) - cos (X-10)
20.10. It's the angle
y=sin(x+20)-cos(x-10)
=sin(x+20)-sin[90-(x-10)]
=sin(x+20)-sin(100-x)
=sin(x+20)-sin[180-(100-x)]
=sin(x+20)-sin(x+80)
Sum difference product
=2cos(x+50)*sin(-30)
=-cos(x+50)
So the range [- 1,1]
The definition field of odd function f (x) is R. when x is greater than 0, f (x) = x ^ 2-2x, find the expression of F (x) on R
Please write the detailed process, thank you
When x = 0, obviously f (x) = 0
Let x0, f (- x) = x & # 178; + 2x
f(x)=-f(-x)=-x²-2x
Therefore, the expression of F (x) on R is a piecewise function
X
|The maximum value of SiNx | + | cos2x |
Must be detailed process, a higher level
It is simple to find the maximum value: when 0 ≤| SiNx | ≤ 1 and x = n π + π / 2, | SiNx | has the maximum value of 1; when 0 ≤| cos2x | ≤ 1 and 2x = m π + π, | cos2x | has the maximum value of 1, that is, x = (M / 2) π + π / 2 | cos2x | has the maximum value of 1; when m / 2 = n, both take the maximum value of 1, therefore, the maximum value of the original function is 2
F (x) = arccosx - π / 4, find the inverse function expression and domain of F [f (x)] why am I wrong?
Inverse function: cos [(arccosx - π / 4) - π / 4]
=cos[arccos(x-π/2)]=cosarccosx=x