0 2 is known, and 3sin 4cos is found in 2 2sin cos 3cos 2 1 is found in cos 2 sin cos

Solution:1(sin^2α+2sinαcosα)/(3cos^2α-1)

sin^2α+2sinαcosα)/(3cos^2α-sin^2a-cos^2a)

sin^2α+2sinαcosα)/(2cos^2α-sin^2a)

tan 2a + 2tana) (2-tan 2a) (the numerator and denominator are divided by cos 2a).

3sin = 4cos i.e. tana = 4 3 brought into the original tense.

sin^2α+2sinαcosα)/(3cos^2α-1)=20

Let sina = 4k and cosa = 3k then (4k) 2 + (3k) 2 = 1 solution gives k = 1 5 or -1 5

So sina = 4 5 and cosa = 3 5

cos^2α+sinαcosα=(3/5)^2+(4/5)*(3/5)=21/25

Hello! The analysis is loud: first analyze the known condition sin +sin 2 =1 given by the problem, you can get sin =cos 2, and then Sun Xiangxing substitutes 3cos 2 +cos 4 -2sin +1 to obtain the result directly

Answer: Solution: by the meaning of the question sin +sin 2 = 1;

This can be obtained: sin = 1-sin 2 = cos 2 , so the original formula = 3 sin + sin 2 -2 sin +1=sin +1-cos 2 +1=sin -sin +2=2

sinθ+cosθ=2sinα

So the two sides are squared.

1+2sinθcosθ=4sin²α

again sin *cos = sin

So 1+2sin = 4sin

1+1-cos2β=2(1-cos2α)

cos2β=2cos2α

Both sides are squared at the same time.

4cos²2α=cos²2β

Original line block = ( 3cos40-sin40))(3cos40+40) sin flat file raid Hu Fang 40cos Ping Zen formula 40

16 (3cos40 2-sin40 2) (cos40 2 + sin40 2) sin square 80

16sin20cos10 cossquare 10

16*2sin10cos10/cos10

32sin10

Hello! The analysis is loud: first analyze the known condition sin +sin 2 =1 given by the problem, you can get sin =cos 2, and then Sun Xiangxing substitutes 3cos 2 +cos 4 -2sin +1 to obtain the result directly

Answer: Solution: by the meaning of the question sin +sin 2 = 1;

This can be obtained: sin = 1-sin 2 = cos 2 , so the original formula = 3 sin + sin 2 -2 sin +1=sin +1-cos 2 +1=sin -sin +2=2

It is obtained from the formula sin 2@+cos 2@=1=sin@+sin 2@, sin@=cos 2@, and the original formula 3sin@+sin 2@-2sin@+1=sin@+sin 2@+1=1+1=2

I wish you progress in your studies! (I use @ to indicate the corner, it's okay, it's funny).