High trigonometry cos 4 x 3 5 7 12 x 7 4 So why sin2x cos 2 2x 7 25. While

Because cos( 2+2x)=cos[2( 4+x)] so the 2 times angle formula.

cos( 2+2x)=2cos 2( 4+x)-1sin2x=-cos( 2+2x) =7 25 There is also the formula sin2x=2sinxcosx

sin2x should be equal to 2sinxcosx

by cos(4+x)=35Open: (2 2)cosx-(2 2)sinx=3 5

cosx-sinx=3√2/5

The square of both sides: 1-2sinxcosx=18 252sinxcosx=7 25 is the answer to sin2x.

tanαcos^3α)/1-sinα)

sinα/cosα)cosα*(1-sin^2α)/1-sinα)sinα*(1+sinα)

Because the stool is buried cos =-3 5, then sin 2 =1-cos 2 =16 25 if 2 "Brother Cong then sin =4 5 original jujube ant = sin 2 +sin =16 25 25 + 4 5=36 25 if 3 2, then sin =-4 5 original = sin 2 +sin =16 25-4 5=-4 25

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

There is still a problem with the title, it should be "cos +cos +2cos cos = 1 2" I lost the root number 2???

cos²α+cos²β+cosαcosβ

1+cos2 ) 2+(1+cos2) 2+cos cos (powered).

1 + cos2α+cos2β)/2 + cosαcosβ

1 + 2 + cos cos (obtained from the above equation "sum difference product").

1 + cos(α+cos(α-cosαcosβ

1 - 2 2)cos( -2cos cos (cos( += - 2 2 substitution)

1 - 2 2) (cos cos + sin sin) 2cos cos (cosine formula with the sum of two angles).

1 + 2/2)(cosα*cosβ-sinα*sinβ)

1 + 2 2) cos ( + invert the cosine formula of the difference between the two angles).

1-1 2 ( cos( +=- 2 2 substitution)

Wrong question, right?

For example, if you want = =3 8, then: cos +cos +cos cos =3cos = is not true.

This question is wrong, there are a lot of things missing, it is known that it should be + =3 4 , and the proof should be cos +cos +2 cos cos cos =1 2

It can be proved that the product of the first two items after the reduction and the difference between the product and the product of the latter item can be proved after offsetting.

If there is a problem with this question, please send me the correct one.

(1) The value of the required formula can be obtained by turning the required formula into -sin2x, and then using the known conditions to obtain the value of sin2x by using the double angle formula.

2) Convert the required formula to sin2x tan(4+x), and find the values of sin(4+x) and cos(4+x) according to the range of x, then the value of tan(4+x) can be obtained, and the value of (sin2x+2sin2x) (1-tanx) can be obtained

cos2( 4+x)=cos( 2+2x)=-sin2x, and cos2( 4+x)=2cos( 4+x)-1=2 9 25-1=-7 25, sin2x=7 25

sin2x+2sin²x)/(1-tanx)

sin2x(1+sinx/cosx)/(1-tanx) →sin2x·sinx/cosx=2sinxcosx·sinx/cosx=2sin²x

sin2x(1+tanx)/(1-tanx)

sin2xtan(π/4+x)．

5π/4＜x＜7π/4，3π/2＜x+π/4＜2π，sin(π/4+x)=√(1-cos²(π/4+x)=-4/5，tan(π/4+x)=sin(π/4+x)/cos(π/4+x)=-4/3．

sin2x+2sin2x)/(1-tanx)

sin2xtan(π/4+x)