Verification: cos ^ 8x sin ^ 8x-cos2x = 1 / 8 (cos6x - cos2x) And CSC ^ 4a (1-cos ^ 4a) - 2cot ^ 2A

Verification: cos ^ 8x sin ^ 8x-cos2x = 1 / 8 (cos6x - cos2x) And CSC ^ 4a (1-cos ^ 4a) - 2cot ^ 2A

cos^8x-sin^8x-cos2x
=(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x
=(cos^4x+sin^4x)*1*cos2x-cos2x
=[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2-1]cos2x
=-(2sinxcosx)^2/2*cos2x
=-sin^2 2x/2*co2x
=-(1+cos4x)*cos2x/4
1/8(cos6x - cos2x)
=1/8*(-2)*sin(8x/2)sin(4x/2)
=-1/4sin4xsin2x
=-1/4*(2sin2xcos2x)sin2x
=-1/4*2(sin2x)^2cos2x
=-1/4(1+cos4x)cos2x
So cos ^ 8x sin ^ 8x cos2x = 1 / 8 (cos6x - cos2x)

It was proved that cos ^ 8 (x) - Sin ^ 8 (x) + (1 / 4) sin2xsin4x = cos2x Note: 8 on COS and sin at the beginning is the eighth power, and X is not on 8, but on COS and sin

cos^8(x) - sin^8(x) + (1/4)sin2xsin4x = [cos^4(x) + sin^4(x)]*[cos^4(x) - sin^4(x)] + (1/4)*sin2x * 2*sin2x*cos2x=[cos^4(x) + 2*cos^2(x)*sin^2(x) + sin^4(x) - 2*cos^2(x)*sin^2(x)] * [cos^2(x) + sin^2(...

cos2x=3/5.sin^4+cos^4x=?

(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2
=1-(1/2)[sin(2x)]^2
=1-(1/2)[1-cos(2x)^2]
=1-(1/2)[1-(3/5)^2]
=1-8/25
=17/25

If cos2x = 3 / 5. What is sin ^ 4 + cos ^ 4x = then

(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2
=1-(1/2)[sin(2x)]^2
=1-(1/2)[1-cos(2x)^2]
=1-(1/2)[1-(3/5)^2]
=1-8/25
=17/25

Given cos2x = root 2 / 3, the value of sin ^ 4x + cos ^ 4 is? Find detailed process!

y=sin^4x+cos^4x
=sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x
=(sin^2 2x+cos^2 2x)^2-(1/2)(2sinxcosx)^2
=1-(1/2)(sin2x)^2
=1-(1/2)(1-cos^2 2x)
=1-1/2*(1-2/9)
=1-7/18
=11/18

Is cosx ^ 2 = 1 + cosx / 2 or equal to 1 + cos2x / 2? Is there still sin2x = 1-cosx?

The first formula is equal to (1 + cos2x) / 2
Sin2x = 1-cosx is wrong, it should be sin2x = 2sinxcosx

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

tan(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)sin(x/2)]/[cos(3x/2)cos(x/2)]
=Sin (3x / 2-x / 2] / [(1 / 2) (cos2x + cosx) (product sum difference)
=2sinx/(cosx+cos2x)
So the original form is established

It is proved that Tan [3x / 2] - Tan [x / 2] = 2sinx / [cosx + cos2x]

Tan (3x / 2) - Tan (x / 2) = sin (3x / 2) / cos (3x / 2) - sin (x / 2) / cos (x / 2) (general divide) = [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 x + cos2x) so the original formula is set up, so the original formula is set up, so the original formula is set up, so the original formula is set up, so the original formula is set up, so the original formula is set up, so the original formula is set up, so the... "

tan(3x/2)-tan(x/2)-2sinx/(cosx+cos(2x))=?

tan(3x/2)-tan(x/2)-2sinx/(cosx+cos2x)
=sin(3x/2)/cos(3x/2)-sin(x/2)/cos(x/2)-2sinx/(cosx+cos2x)
=[sin(3x/2)cos(x/2)-cos(3x/2)sin(x/2)]/cos(3x/2)cos(x/2)-2sinx/(cosx+cos2x)
=2sin(3x/2-x/2)/[cos(3x/2+x/2)+cos(3x/2-x/2)]-2sinx/(cosx+cos2x)
=2sinx/(cos2x+cosx)-2sinx/[cosx+cos(2x)]
=0

Sin2x / (1-cos2x) - (1 / TaNx) simplification

sin2x/(1-cos2x)-(1/tanx)
=2sinxcosx/(2sin²x）-cosx/sinx
=cosx/sinx-cosx/sinx
=0