Prove the following identities: 1 2sin 2 x cos 2 x cos 2cos 2x 1 sin sin 2x .

You're a girl! Simple trigonometric transformations, the sine and cosine theorem will not. You didn't learn, or did the teacher not teach you? If you are an art and sports student, ask for a teacher's help.

I'm in a bad mood and don't want you. (It's "ice", not "pawns"; Don't give one full name, it's too insincere! )

Certificate: 1)(cosx-1) +sin x

cos²x-2cosx+1+sin²x

cos²x+sin²x)+1-2cosx2-2cosx

2)1+tan²x

1+sin²x/cos²x

cos²x+sin²x)/cos²x

1/cos²x

sinx/(cos²xsinx)

sinx/cosx)/(sinxcosx)tanx/(sinxcosx)

3)sin²x+sin²xcos²x+cos²x(sin²x+cos²x)+sin²xcos²x1+sin²xcos²x

Is there a mistake in this question? 】,1,Xinxin's cat report.

sin 4x+sin xcos x+cos x=1 sin 4x+sin xcos x+cos x =sin x(sin x+cos x)+cos x =sin x+cos x =1,:(1)(cosx-1) +sin x=2-2cosxsin 2x+xos 2x=1 Answer: Bucket oak comes out (sin 2x) the square of the table sinx.

3) It is miscognizable.

2) Can you also be detailed1: (cosx-1) =cos 2x-2cosx+1 substitution (cosx-1) +sin x=2-2cosx 3:sin x+sin xcos&.

0, which proves the following identity: (1)(cosx-1) near the empty sare + sin x=2-2cosx

2)1+tan x=tanx (sinxcosx)3)sin x+sin xcos then which x+cos x=1

Apparently sin( +sin, and cos( -cos is known by the formula next to the slag.

2sin(π+cos(π-

2(-sinα)*cosα)

2sinα*cosα

sin2α

First, reduce to 2xcosx0

In addition, f(x)=tanx+sinx-2x, the agitator knows that it is monotonically increasing at 0 x (muffled 2).

Method: You can use the unit circle to determine x as the acute angle of the slag stove, and then judge it according to the positive and negative conditions of sinx and cosx in each quadrant (this high school textbook has details).

It can also be directly decomposed with the differential product. (Proven below with the sum of the accumulation).

Only cos(x+2 )cosxcos 2-sinxsin 2=-sinx

Similarly: sin( -x)=sin cosx-cos sinx=sinx,8, prov:

1）cos(x+2/π)sin[2/π-x+2/π)sin(-x)=-sinx

2)sin( -x)=sin[ -x)]=sinx,0, "1" cos(x+2 )cos cos2 e.g. sin2 sin =0*cos 1*sin = sin

2》∵sin(π-x)=sinπcosχ－sinχcosπ=0*cosχ－（1）*sinχ=sinχ,0,

Proof of Credit: False Reputation.

1)cos(x+2 )sin[2 -x+2 dust reputation burning )]sin(-x)=-sinx

2)sin( -x)=sin[ -x)]=sinx

Prove. 1-2sinacosa) (sin a-cos a)(sina-cosa) sina + cosa) (sina-cosa) (sina-cosa) (sina+cosa) up and down at the same time divide by sina, (1-tana) (1+tana).

, left=cos +sin +sin (1+tan )=1+sin cos

1+tan^α

1 cos = right.

2.Left = (2cos + sin) (cos + 2sin ) = 2cos + 5sin cos + 2sin = 2 + 5 sin cos = right.

3.Left = (1 cos a) (1 sin a) = (sina cosa) = tan a, right =

-sina/cosa)^

tan a = left.

4.Left = (tana-tanb) [1 tana-1 tanb]=-tanatanb, right = tanatanb≠ left.

Please check question 4.