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If cosx + cosy = 1 / 2, SiNx siny = 1 / 3, then cos (x + y) =?
There is cos (x + y) = cosxcosy sinxsiny, and the square of the formula given in the title is added to the left and right sides of the formula as follows: cos ^ 2x + cos ^ 2Y + 2cosxcso y + sin ^ 2x + sin ^ 2y-2sinxsiny = 13 / 36. From cos ^ 2x + sin ^ 2x = 1 cos ^ 2Y + sin ^ 2Y = 1, we can get 1 + 2cosxcso oy + 1-2sinxsiny
If SiNx + siny = 0.4, cosx + cosy = 1.2, then cos (X-Y)=______ .
∵sinx+siny=0.4，①
cosx+cosy=1.2，②
① 2 + 2 + 2 2, 2 + 2 SiN x sin y + 2 cosx cosy = 1.6,
∴cos（x-y）=-1
5，
So the answer is: - 1
5．
SiNx siny = - √ 3 / 3 cosx cosy = - 1 / 3 then cos (X-Y)
sinx-siny=-√3/3
Square both sides at the same time
sin²x-2sinxsiny+sin²y=1/3 ①
cosx-cosy=-1/3
Square both sides at the same time
cos²x-2cosxcosy+cos²y=1/9 ②
From ① + ②, we can get the following results:
-2sinxsiny-2cosxcosy=4/9
That is - 2 (sinxsiny + cosxcosy) = 4 / 9
-2cos(x-y)=4/9
cos(x-y)=-2/9
Answer: - 2 / 9
Find the function y = 2cos (x + π 4)cos(x−π 4) + The range and minimum positive period of 3sin2x
y＝2cos(x+π
4)cos(x−π
4) +
3sin2x
=2(1
2cos2x−1
2sin2x)+
3sin2x
=cos2x+
3sin2x
=2sin(2x+π
6)
The function y = 2cos (x + π)
4)cos(x−π
4) +
The value range of 3sin2x is [- 2,2],
The minimum positive period is π;
Let f (x) = the radical sign 3sin2x + 2cos ^ 2x + 2, and find the minimum positive period sum range of F (x)
Because f (x) = √ 3sin2x + 2 (cosx) 2 + 2
=√3sin2x+1+cos2x+2
=2(√3/2 * sin2x + ½ * cos2x)+3
=2(cosπ/6 * sin2x + sinπ/6 * cos2x)+3
=2sin(2x+π/6)+3
So the minimum positive period of F (x) is 2 π / 2 = π
Because f (x) | max = 2 + 3 = 5, f (x) | min = - 2 + 3 = 1,
So the range of F (x) is [1,5]
Let (2x) = (2x) = (2x) = (2x); Seeking monotone decreasing interval of F (x) by ＜ 2 ＞
f(x)=√3 sin2x+cos2x+3=2sin(2x+π/6)+3
1) The minimum positive period is π
Range: [1,5]
2) Monotone decreasing interval: 2K π + π / 2
The known function f (x) = 2cos ^ 2x + radical 3sin2x-1 (1) Find the monotone increasing interval of F (x) (2). When x takes what value, f (x) takes the maximum value? What is the maximum value?
The same is the problem of simplification first
f(x)=2cos^2x+√3*sin2x-1
＝2(cosx)^2-1+√3*sin2x
=cos2x+√3*sin2x
=2(1/2*cos2x+√3/2*sin2x)
=2(sinπ/6*cos2x+cosπ/6*sin2x)
=2sin(2x＋π/6)
Because the monotone increasing interval of SiNx is (2k π - π / 2,2k π + π / 2)
Therefore, the solution of monotone increasing interval of this function is as follows:
2kπ-π/2 < 2x+π/6 < 2kπ+π/2
obtain
kπ-π/3 < x < kπ+π/6
The reason why it becomes K π here is that the period of this function is no longer 2 π, but π
The interval representation method is: (K π - π / 3, K π + π / 6)
(2) Generally, for SiNx, when x = 2K π + π / 2, the maximum value is 1,
Substituting into this function
2x+π/6=2kπ+π/2
x=kπ+π/6
When x = k π + π / 6, f (x) has a maximum value of 2
To correct your previous question, "shift the image of the function f (x) = cosx + SiNx to the left by m, (M > 0) units, and the corresponding function of the image is even function. Find the minimum value of M"
Because I haven't done it for many years, I remember a wrong formula. After the function is simplified, it should be = √ 2 sin (x + π / 4). At the beginning, I mistakenly wrote it as minus. I didn't want to mistake people. I wanted to help you. Please don't mislead you. So I hope you can revise it by yourself if you see it. I'm really sorry
If θ∈ [- π] 12，π 12] Then the function y = cos (θ + π) 4) The minimum value of + sin2 θ is () A. 0 B. 1 C. 9 Eight D. Three 2−1 Two
Y = cos (θ + π 4) + sin2, θ = cos (θ + π 4) - cos (2 θ + π 2) = - Cos2 (θ + π 4) + cos (θ + π 4) = - 2cos2 (θ + π 4) + cos (θ + π 4) + 1 ∵ θ ∈ [- π 12, π 12], ν12 ≤ cos (θ + π 4) ≤ 32, so let cos (θ + π 4) = ty = - 2 (t-14) 2 + 98
Given the function f (x) = min {SiN x, cos x}, then the value range of F (x) is [- 1, root 2 / 2] Tomorrow will be the exam, please come back quickly! I don't know why I got the root 2 / 2? Why not 1?
Because the smaller one of SiNx and cosx is found, when both are greater than 0, SiNx increases and cosx decreases, and vice versa. Therefore, when the two values are the same, the minimum value between them is the largest, which is calculated as √ 2 / 2
Find the value range of SiN x of the function y = 3 times the root sign divided by 2 minus cos X
y=√7/2[sinx*√21/7-cosx*2√7/7)
Let cos θ = √ 21 / 7, sin θ = 2 √ 7 / 7),
y=√7/2sin(x-θ),
Range: y ∈ [- √ 7 / 2, √ 7 / 2]
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