Let the function fx vector a vector b vector c where vector a

Suppose function fx vector a vector b vector c where vector a sinx

Suppose vector d(h,k)

So x'=x-h; y'=y-k

x=x'-h ; ​​y=y'-k

Then put the above formula into the original F(x)

Get y'-h=2 √2sin(2x-2h-3π/4)

Now see the condition of "making the image obtained after translation symmetrical about the center of the coordinate origin" in the question

That is to say, when x=0, the equation after the first translation g(0)=0

So at this time -2h-3π/4=kπ

’s h=3π/8-kπ/2

Then we get d(3π/8-kπ/2,-2)

The key to answering this question is to set up the vector according to the translation method

This x'=x-h; y'=y-k

x=x'-h ; ​​y=y'-k

f(x)=vector a*(b c)

From the question f(x)=(sinx,-cosx)*(sinx-cosx,-3cosx sinx)

f(x)=sinx(sinx-cosx)-cosx(-3cosx sinx)

=sinxsinx-sinxcosx 3cosxcosx-sinxcosx

=sinxsins 3cosxcosx-2sinxcosx

=sinxsinx cosxcosx 2cosxcosx-2sinxcosx

=cos2x-sin2x

=square root 2/2 sin(2x 45 degrees)

Suppose function fx vector a vector b c where vector a sinx cosx vector b sinx

(1)f(x)=a(b c)=ab ac=sinxsinx 3coxcox-2sinxcosx

=2cosxcosx-sin2x 1

=-sin2x cos2x 2

=√2sin(2x 3π/4) 2

(2) When x belongs to [3π/8, 7π/8], 2x 3π/4 belongs to [3π/2, 5π/2]

According to the properties of sinx, f(x) is monotonically increasing in [3π/8,7π/8]

(3) First translate y=cosx to the right by π/2 units to get y=cos(x-π/2)=sinx

If x remains unchanged and y increases by √2 times, we get y=√2sinx

If y remains unchanged, x is reduced to 1/2 of the original value, and y=√2sin(2x)

Move 3π/8 units to the left to get y=√2sin(2x 3π/4)

Finally translate upward by 2 units to get y=√2sin(2x 3π/4) 2

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