In order to get the image of the function y = sin 2x, we just need to transform the function y = cos (2x - π) 4) Image direction of______ Translation π Eight units is enough

In order to get the image of the function y = sin 2x, we just need to transform the function y = cos (2x - π) 4) Image direction of______ Translation π Eight units is enough

∵y=sin 2x=cos（2x-π
2），
∴y=cos （2x-π
4) Image shift to the right π
8 units
y=cos[2（x-π
8）-π
4）]=cos（2x-π
2），
So the answer is: right

To get the image of the function y = cos (2x - π / 4), just shift the image of function y = sin2x to the right by π / 8 units. How to get it? If my solution is right or wrong, what's wrong? Y = cos (2x - π / 4) is equivalent to y = sin (2x-3 π / 4), so just shift y = sin2x to the right by 3 π / 8 units

For this kind of problem, moving the image to the left and right is actually to add and subtract X in the function from the left to the right
Shift the function y = sin2x to the right π / 8 units,
That is: y = sin2 (x - π / 8) = sin (2x - π / 4)

In order to get the image of function y = cos (2x + Pai / 4), how many units can the image of function y = sin2x be shifted to the right at least?

5/8π

Proof (COS ^ 2x sin ^ 2x) (COS ^ 4x + sin ^ 4x) + 1 / 4sin2xsin4x = cos2x

It is very simple, mainly to simplify 1 / 4 * sin2x * sin4x: 1 / 4 * sin2x * 2 * sin2x * cos2x = 2Sin ^ 2x * cos ^ 2x * cos2x;
Then the right formula is multiplied by cos ^ 2x sin ^ 2x to cos2x, and the common factor is extracted. The result is clear at a glance

Proof! Cos ^ 8x sin ^ 8x + 1 / 4sin2xsin4x = cos2x Ask the great God of Mathematics for help. Thank you

cos^8x-sin^8x+1/4sin2xsin4x-cos2x
=(cos^4x+sin^4x)(cos^4x-sin^4x)+1/2sin²2xcos2x-cos2x
=(cos^4x+sin^4x)(cos²x-sin²x)+2sin²xcos²x(cos²x-sin²x)-cos2x
=cos^6x-sin^6x+sin²xcos^4x-sin^4xcos²x-cos2x
=cos^4x-sin^4x-cos2x
=cos²x-sin²x-cos2x
=cos2x-cos2x
=0
So cos ^ 8x sin ^ 8x + 1 / 4sin2xsin4x-cos2x = 0
cos^8x-sin^8x+1/4sin2xsin4x =cos2x

If cos2x = a, then the value of sin ^ 4x + cos ^ 4x is equal to?

sin^4x+cos^4x=(1-cos2x)^2/4+(1+cos2x)^2/4
=(1-a)^2/4+(1+a)^2/4
=(1+a^2)/2

If cos2x = √ 2 / 3, how much is cos ^ 4x + sin ^ 4x?

cos^4x+sin^4x
=[(1+cos²2x)/2]^2+[(1-cos²2x）/2]^2
Then we bring in cos2x = √ 2 / 3,
Get 11 / 18
I made a mistake = =. In the final... Careless... Wrong symbol

Verify (sin2x) ^ 2 + (2 * (cosx) ^ 2) * cos2x = 2 (cosx) ^ 2

(sin2x)^2+(2*(cosx)^2)*cos2x
=(2sinxcosx)^2+2*(cosx)^2*(1-2(sinx)^2)
=2*(cosx)^2*2(sinx)^2+2*(cosx)^2*(1-2(sinx)^2)
=2*(cosx)^2*[2(sinx)^2+(1-2(sinx)^2)]
=2(cosx)^2

Prove SiNx + sin3x + sin2x = 1 + cos2x + cosx

I have a question, is this proposition true? Left = SiNx + 2sinxcosx + 3sinx (cosx) ^ 2 - (SiNx) ^ 3 = 4sinx (cosx) ^ 2 + 2sinxcosx = 2sinxcosx (2cosx + 1) = 2sinx (1 + cos2x + cosx) 2sinx * right. Then it is equal only when 2sinx = 1. Is it contradictory

It was proved that Tan (3x / 2) - Tan (x / 2) = (2sinx) / (cosx + cos2x)

(3x / 2) - Tan (x / 2) = sin (3x / 2) / cos (3x / 2) - sin (x / 2) / cos (x / 2) (General Division) = [sin (3x / 2) cos (x / 2) - cos (3x / 2) cos (x / 2) - cos (3x / 2) sin (x / 2)] / [cos (3x / 2) cos (x / 2)] = sin (3x / 2-x / 2] / [(1 / 2) (cos2x + cos x) (the product and difference) = 2sinx / (COS x + cos 2x) 2 SiNx / (COS x + cos 2x) therefore, the original formula of the original formula is the original formula, therefore, the original formula of the original formula is the original formula of the original formula, the original formula of the original formula, so the original formula of the original formula is the original set up