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The minimum value of the function f (x) is - 2 √ 2 when f (x) = asinx + bcosx (a > 0), f (3 / 4 π) = 2 √ 2 (1) Find the value of a and B (2) the value of F (π / 3) (3) when f (x) = 0, find the value of X
(1) Because the minimum value of the function f (x) is - 2 √ 2, a squared + B squared = 8
Substitution shows that f (3 / 4 π) = (√ 2) * (a-b) / 2 = 2 √ 2
So A-B = 4
a=2,b=-2
(2) Just take it in, f (π / 3) = √ 3-1
(3) When f (x) = 0, 2 √ 2Sin (x - π / 4) = 0, X - π / 4 = k π, x = k π + π / 4
If f (x) = asinx + bcosx gets the minimum value at x = n / 4, then, what function is f (x) and what symmetry is f (x) when y = (3N / 4-x)?
The expression is not very clear. N should be "Pai"
Auxiliary angle formula asinx + bcosx = (√ a ^ 2 + B ^ 2) sin (x + θ)
X = n / 4 gets the minimum value, so θ can be taken as (3N / 2) - N / 4 = 5N / 4
When y = (3N / 4-x), f (x) = (√ a ^ 2 + B ^ 2) sin (3N / 4-x + 5N / 4) = - (√ a ^ 2 + B ^ 2) sin (x)
This is an odd function, symmetric about the origin [and (KN, 0)]
The image passing through points (π / 3,0) and (π / 2,1) of the given function f (x) = asinx + bcosx 1: Find the value of real numbers a and B 2: When x is the value, f (x) gets the maximum value
1. Substituting: F (π / 3) = the root of 2, 3A + B / 2
f（π/2）=a+0=1
A = 1 b = - radical 3
2. F (x) = SiNx radical 3cosx
=2sin(x-π/3)
When sin (x - π / 3) = 1, that is, X - π / 3 = π / 2 + 2K π, the maximum value of F (x) is obtained
In this case, x = 5 π / 6 + 2K π
Maybe the calculation is wrong, but the method is like this~
The image of the function f (x) = asinx + bcosx passes through points (π / 3,0) and (π / 2,1) (1) Find the values of functions a and B; (2) When x is the value, f (x) gets the maximum value?
Substitution
A=1
B = - radical 3
F (x) = 2Sin (X-60 '), X-60' = 90 '+ 360 ` K, x = 150' + 360 ` K, K belongs to the set of integers
14. Given the function f (x) = 1 / 2sinx - the square of root 3sin X / 2 + radical 3 / 2 + 1 (1) the minimum positive period and the maximum value of the solution; (2) the monotone increasing interval
(Ⅰ)f（x）=½sinx-√3/2（1-cosx）+√3/2+1
=½sinx+√3/2cosx+1
=sin（x+π/3）+1
The minimum positive period T = 2 π / ω = 2 π
Max = 1 + 1 = 2
(II) from 2K π - π / 2 ≤ x + π / 3 ≤ 2K π + π / 2
The monotonic increasing interval is [2K π - 5 π / 6,2k π + π / 6] (K ∈ z)
Let f (x) = a * B, where vector a = (2Sin (π / 4 + x), cos2x), B = (sin (π / 4 + x), - root 3), X belongs to R, Find: 1. Monotone increasing interval of function f (x) 2. When x belongs to [0, π / 2], find the value range of function f (x)
(x) = a * b = 2Sin (x + π / 4) cos (x + π / 4) cos (x + π / 4) cos (x + π / 4) - √ 3cos 2x = sin (2x + π / 2) - √ 3cos 2x = (1 - √ 3) cos2x1. The single increase interval is 2x ∈ [2K π - π / 2,2k π + π / 2] x ∈ [K π - π / 4, K π + π / 4] 2.x belongs to [0, π / 2], 2x ∈ [0, π] f (x) f (x) f (x) x ∈ [0, π [0, π] f (x) f (x) f (x) f (x) f (x the range of values is [1 - √ 3, √ 3-1]
Let f (x) = 2Sin 2 (π / 4 + x) - (√ 2) cos2x-1, X ∈ R 1. Find the minimum value of F (x) Let f (x) = 2Sin 2 (π / 4 + x) - (√ 2) cos2x-1, X ∈ R 1. Find the minimum value of F (x) 2. Let P: X ∈ [π / 4, π / 2], Q: | f (x) - M|
(1)f(x)=2sin²(π/4+x)-（√2）cos2x-1=2*(1-cos²(π/4+x))-（√2）cos2x-1=2[1-(cos(π/2+2x)+1)/2]-（√2）cos2x-1=1+sin(2x)-（√2）cos2x-1=sin(2x)-（√2）cos2x=√3[(1/√3)sin(2x)-（√2/√3）cos2...
Given the function f (x) = (sin2x cos2x + 1) / 2sinx, find the definition domain of F (x)
{x|x≠kπ,k∈Z}
Given the function f (x) = (sin2x + cos2x + 1) / 2cos2x, find the function definition domain and value domain
As long as cos2x ≠ 0, the domain is defined as {x} x ∈ R, and X ≠ K Π + Π / 2, K ∈ Z}
f（x）=1/2（tan2x+1/cos2x+1）
The range of Tan 2x and COS 2x in the first image limit is (0, + ∞)
The range of tan2x and cos2x in the third image limit is (- ∞, 0)
When x = k Π / 2 + Π / 2, f (x) = 0
So the range of F (x) is (- ∞, + ∞)
Given the function f (x) = cos2x / sin (x + Pai / 4), if f (x) = 4 / 3, find sin2x
cos2x/sin(x+π/4)＝cos^2x-sin^2x/(√2/2（sinx+cosx))＝√2（cosx-sinx）＝4/3
Square the two sides to get:
2（1－sin2x)=16/9
The solution is: sin2x = 1 / 9
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