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On line equal known vector M = (cosx, SiNx), n = (2 radical 2 + SiNx, Given the vector M = (cosx, SiNx), n = (2 radical 2 + SiNx, 2-cosx), where x is in (Π / 2, Π), and Mn = 1. (1) find the value of sin (x + Π / 4); (2) find the value of COS (x + 7 Π / 12)
(1) The vector M = (cosx, SiNx) vector n = (2 √ 2 + SiNx, 2 √ 2-cosx) has the following vector n = (2 √ 2 + SiNx, 2 √ 2-cosx), Mn = cosx (2 √ 2 + 2 + SiNx) + SiNx (2 √ 2-cosx) = 1sinx + cosx = 2 / 4sinx × √ 2 / 4sinx × × √ 2 / 2 + cos × × ×, 2 / 2 = (2 / 4) × (2 / 2) sinxcos π / 4 + cosxsin π / 4 = 1 / 4 (4) 1 / 4 (x) SiNx (x + π / π / π / π / π / 4 = 1 / 4) = 1 /
The minimum positive period of the function FX = 1 + SiNx / cosx is?
=1+tanx
The minimum positive period is π
Proof of periodic function by FX = 4 | SiNx | cosx
f(x)=4|sinx|cosx
=4|sin(x+2π)|cos(x+2π)=f(x+2π)
The domain of F (x) = 4 | SiNx | cosx is a real number field, and X + 2 π is also in the domain
F (x) periodic function
The value range of the function y = 3sin ^ 2 x - 2 root sign 3sin x cos x + 5cos ^ 2 x on "0,45" is
Y = 3sin ^ 2 x - 2 radical 3sin x cos x + 5cos ^ 2 x = 4-cos2x - radical 3sin2x = 4-2cos (2x + 60 degrees)
So it belongs to the (60,4) root field
The value range of the function f (x) = SiNx / radical (5 + 4cosx) [0 < = x < = 2 π] is
First of all, this is a continuous differentiable function. By finding the boundary value and the extreme value, we can obtain the value range derivation f '(x) = cosx / root sign (5 + 4cosx) + 2 (SiNx) ^ 2 / (5 + 4cosx) ^ (3 / 2) = [cosx (5 + 4cosx) + 2 (SiNx) ^ 2] / (5 + 4cosx) ^ (3 / 2) = [2 (cosx) ^ 2 + 5cosx + 2] / (5 + 4cosx) ^ (3
The range of y = SiNx / Radix (5 + 4cosx) Don't copy that on the Internet
y^2=sin^2(x)/(5+4cosx),
Let t = (5 + 4cosx),
∵cosx∈[-1,1],∴t∈[1,9]
Then cosx = (T-5) / 4,
sin^2(x)=1- cos^2(x)=1-(t-5)^2/16,
y^2=sin^2(x)/(5+4cosx)=[ 1-(t-5)^2/16]/t,
16y^2=[ 16-(t-5)^2]/t,
16y^2=(-t^2+10t-9)/t,
16y^2=10-(t+9/t))
Because the graph of the function T + 9 / T is a "√", it decreases on (1,3) and increases on (3, + ∞),
When t ∈ [1,9], the minimum value of the function T + 9 / T is 6 (obtained when t = 3), and the maximum value is 10 (taken when t = 1 or 9)
Therefore, 10 - (T + 9 / T)) ∈ [0,4]
That is, 16y ^ 2 ∈ [0,4]
y∈[-1/2,1/2].
The range of y = 2cosx + sinx-1 is [- 2,9 / 8],
y=2cos²x＋sinx－1
=2(1－sin²x)＋sinx－1
=－2(sinx－1/4)＋9/8
When SiNx = 1 / 4, the maximum value is 9 / 8; when SiNx = - 1, the minimum value is - 2
Y = (sinx-2) / (2cosx-1)
y=（sinx-2)/(2cosx-1)
sinx-2=y(2cosx-1)
sinx-2=2ycosx-y
2ycosx-sinx=y-2
√(1+4y^2)cos(x+θ)=y-2
∵√(1+4y^2)cos(x+θ)=y-2
∴- √(1+4y^2)
The range of y = SiNx + 2cosx + 1
y=sinx+2cosx＋1
=√5sin(x+a)+1
therefore
The value range is [- √ 5 + 1, √ 5 + 1]
Find the range of y = (2cosx SiNx + 2) / (cosx + sinx-1)
If you don't understand, please tell me your email and let me send you the process that people can understand. Y = (2cosx SiNx + 2) / (cosx + sinx-1) (cosx + sinx-1) y = 2cosx SiNx + 2 (Y-2) cosx + (y + 1) SiNx = y + 2sqrt [(Y-2) ^ 2 + (y + 1)... "
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