Knowing that sin 2x cos 2x 1 5, then tan 2x

sin 2x-cos 2x=1 5 can be written as (sin 2x-cos 2x) 1=1 5

sin^2x-cos^2x）/1=1/5

sin 2x-cos 2x) (sin 2x+cos 2x)=1 5sin 2x-cos 2x) (sin 2x+cos 2x)=1 5 divide the left top and bottom by cos 2x

In this way, you can figure it out, and you can calculate the rest yourself.

sinx+cosx）²=1/25

2sinxcosx=1-1/25=-24/25sinx-cosx）²=1-2sinxcosx=49/25-π/2sinx-cosx=-7/5②

sinx+cosx=1/5①

Gain: Reputation limb sinx=-3 repentance digging 5

Debi core: cosx = 4 5

tanx=sinx/cosx=-3/4

Let's give you the process of solving the problem.

Because sinx+cosx=1 3

The square gives 1+2sinxcosx equal to 1 9

sinxcosx is equal to -4 9

sinx-cosx) squared = 1 + 8 9 = 17 9 because - 2, so cosx-sinx is greater than zero equal to the root number of thirds of the seventeenth, then the root number of the ninth is seventeen.

That's right, there is a root number.

cos( 4+x)=3 5, 3 2 4+x 2 , so sin( 4+x)=-4 5, sin2x=-cos( 2+2x).

cos[2(π/4+x)]

[1-2sin²(π4+x)]=7/25.

sinx+cosx=√2* sin（π/4+x)=-4√2/5,cosx –sinx=√2* cos（π/4+x)=3√2/5,(sin2x+2(sinx)^2)/(1-tanx)

2sinx(sinx+cosx)]/1-sinx/cosx)……Multiply the numerator and denominator by cosx to get the following equation.

2sinx cosx (sinx+cosx)]/cosx -sinx)

sin2x (sinx+cosx)]/cosx -sinx)

Expand the angle. From the known obtained, cos(2+2x)=1-2cos 2(4+x), then cosx can be solved, and then sinx can be solved by cos 2+sin 2=1, so that the original formula can be solved. It can be seen that you are in your second or third year of high school, practice solving it yourself, so that you will gradually become proficient, in fact, this kind of problem is the transformation of angles, as long as you memorize these formulas, it will be no problem, of course, you must be proficient.

cos( 4+x)=3 5, 3 2 4+x 2 , so sin( 4+x)=-4 5, sin2x=-cos( 2+2x).

cos[2(π/4+x)]

[1-2sin²(π4+x)]=7/25.

sinx+cosx=√2*sin（π/4+x)=-4√2/5,cosx–sinx=√2*cos（π/4+x)=3√2/5,(sin2x+2(sinx)^2)/(1-tanx)

2sinx(sinx+cosx)]/1-sinx/cosx)……Multiply the numerator and denominator by cosx to get the following equation.

2sinxcosx(sinx+cosx)]/cosx-sinx)=[sin2x(sinx+cosx)]/cosx-sinx)

Solution: masking sinx+cosx=1 5

sinx+cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx=1+2sinxcosx=(1/5)^2=1/25

2sinxcosx=-24/25

Ling Jing (cosx-sinx) 2=(cosx) 2+(sinx) 2-2sinxcosx=1+24 25=49 25

/20，sinx<0

cosx-sinx> macro virtual 0

cosx-sinx=7/5

sin(2x)+2(sinx)^2]/(1-tanx)

2sinxcosx+2(sinx)^2]/(1-sinx/cosx)

2sinx(cosx)^2+2(sinx)^2(cosx)]/cosx-sinx)

2sinxcosx(sinx+cosx)/(cosx-sinx)

From sinx+cosx=1 5: (sinx+cosx) =1+2sinxcosx=1 25, 2sinxcosx=-24 25, so (sinx cosx) =1 2sinxcosx=49 25, so sinx cosx= 7 5, it can be seen that sinx=-3 5, cosx=4 stroop 5 or sinx=4 5, cosx=-3 5

Again- then tanx=-3 4