Given the vector a = (SiNx, cosx), B = (SiNx, SiNx) if x ∈ [- 3 π / 8, π / 4] function f (x) = the maximum value of λ a * B

Given the vector a = (SiNx, cosx), B = (SiNx, SiNx) if x ∈ [- 3 π / 8, π / 4] function f (x) = the maximum value of λ a * B

f(x)=λa*b=λ(sin^2x+sinx*cosx)
=λ[(1-cos2x)/2+sin2x/2]
=(λ/2)[1+sin2x-cos2x]
=(λ/2)[1+√2sin(2x-π/4)]
Because x ∈ [- 3 π / 8, π / 4], so (2x - π / 4) ∈ [- π, π / 4],
Let t (x) = 1 + √ 2Sin (2x - π / 4), so - 1

Let a = (cosx + SiNx, SiNx), B = (cosx + SiNx, - 2sinx), and f (x) = a · B. find the value range of F (x) where x belongs to [0, π / 2]

f(x)=a·b=(cosx+sinx)²-2sin²x
=cos²x+sin²x+2sinxcosx-2sin²x
=1-2sin²x+2sinxcosx
=1-（1-cos2x）+sin2x
=cos2x+sin2x
=√2sin（2x+π/4）
0=

Let f (x) = a · B + 1. Let a = (2sinx, 2sinx), B = (SiNx, cosx) Let f (x) = a · B + 1 when a vector a = (2sinx, 2sinx), B = (SiNx, cosx) 1) Finding the minimum positive period of function f (x) 2) When x ∈ [0, π / 2], find the maximum value of function f (x) and the set of X when obtaining the maximum value

0

Vector a = (cosx + 2sinx, SiNx) vector b = (cosx SiNx, 2cosx) f (x) = vector a * vector b find the monotone interval of F (x)

a=(cosx+2sinx,sinx),b=(cosx-sinx,2cosx)
f(x)=a·b=(cosx+2sinx)(cosx-sinx)+2sinx*cosx
=(cosx)^2+sinxcosx-2(sinx)^2+2sinxcosx
=(cosx)^2-2(sinx)^2+3sinxcosx
=(1+cos2x)/2-2(1-cos2x)/2+(3/2)sin2x
=-1/2-(3/2)cos2x+(3/2)sin2x
=-1/2+(3√2/2)[√2/2)sin2x-(√2/2)cos2x]
=-1/2+(3√2/2)sin(2x-π/4)
The function SiNx is at - π / 2 + 2K π = - π / 8 + K π = - π / 8 + K π = - π / 8 + K π=
Homework help users 2017-11-02
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Given the vector a (COSA, Sina), B (cosx, SiNx), C = (SiNx + 2sinx, cosx + 2cosa), where 0 1.c=(sinx+2xina,cosx+2cosa) 2. A vertical C

There are errors in both vector C and the second question
(1) First of all, vector multiplication yields f (x) = 2sinxcosx + 1.414 (SiNx + cosx) = (SiNx + cosx) 2 + 1.414 (SiNx + cosx) - 1
The last one is to calculate the range, and then it is a quadratic polynomial, and then find the maximum value. There should be no problem. There is a problem in saying. It seems that the formula compiler can not be used, so it can only be like this
(2) A * b = cosacosx + sinasinx = cos (x-a), because a * b = | a | B | cos π / 3
So we get the equation, the main range, so there are two values, x-a = π / 3 or x-a = - π / 3
Similarly, a * C = sin (a + x) + 4sinacosa, a * C = (5 + 4cos (A-X)) 1 / 2 * cos π / 2 = 0
If you can't do it, you can say it

If f (cosx) = cos2x, then the solution set of F (SiNx) = 2sinx-1 is

f(cosx)=cos2x=2(cosx)^2-1
So f (SiNx) = 2 (SiNx) ^ 2-1 = 2sinx-1
sinx=0,1
So x = k Π, or x = Π / 2 + 2K Π (K ∈ z)
(I'm sorry, the answer was wrong the first time)

Let a = (2sinx, - cos2x). Vector b = (6, - 2 + SiNx). Vector C = (cosx, SiNx). Where 0 ≤ x ≤ Pie / 2 1) Let f (x) = a * (B-C) + 3b Λ 2, find the maximum value of F (x)

1) Vector a ‖ vector B, x1y2 = x2y1
2Sinx(-2+Sinx)=6(-2Cos2x)
-4Sinx+2(Sinx)^2=-12[1-2(Sinx)^2]
-4Sinx+2(Sinx)^2+12-24(Sinx)^2=0
22(Sinx)^2+4Sinx-12=0
The positive value x = (v67-1) / 22 is obtained
2)f(x)=a*(b-c)+3b∧2
=(2sinx,-cos2x)*(6-cosx, -2)+3[36+4-4sinx+(sinx)^2]
=2sinx(6-cosx)-cos2x*(-2)+3[36+4-4sinx+(sinx)^2]
Simplify and merge

How to simplify the calculation of velocity f (x) = cos2x / SiNx + cosx + 2sinx

Cos2x = cosx squared - SiNx squared, so you can do it next

It is known that the quadratic function f (x) for any XX has f (1-x) = f (1 + x) a = (SiNx, 2), B = (2sinx, 1 / 2), C = (cos2x, 1) It is known that the quadratic function f (x) has f (1-x) = f (1 + x) for any x R, let vector a = (SiNx, 2), B = (2sinx, 1 / 2), C = (cos2x, 1), d = (1,2). If 0 ≤ x ≤ π, find the solution set of inequality f (a * b) > F (c * d)

a*b=sinx^2+1
c*d=cosx^2+1
When f (x) is an increasing function in x > 1
a*b>c*d
There are SiNx ^ 2 > cosx ^ 2
So when pi / 4 is a minus function
Zero
Job help users: November 10, 2017
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Given the vector M = (cosx + SiNx. √ 3cosx), vector n = (cosx SiNx, 2sinx), Let f (x) = m × n + 1, Ask ① to find the symmetry center of the image of the function f (x). ② if the angle a is the maximum inner angle of the acute triangle, find the value range of F (a)

Because vector M = (√ 3sinx, cosx), vector n = (cosx, cosx), so √ 3sinx / cosx = cosx / cosx, so TaNx = √ 3 / 3, so x = 30 ° so sinxcosx = √ 3 / 4... The function f (x) = vector m · vector n, so f (x) = √ 3sinxcosx cosx = √ 3 / 2sin2x 1 / 2cos2x 1 / 2 =