If Sin23 is known, then cos 2 .

cos( -2 ) = -cos2 = - (square of cos - square of sin).

The square of the sin = 2 the square of 3 = 4 9

Because the square of cos + the square of sin = 1, the square of cos = 1-4 9 = 5 9

So cos( -2 ) = -cos2 = - (square of cos - square of sin) =-(5 9-4 9) = -1 9

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One is , one is a, and it can't be solved.

Did the landlord make a mistake? It's supposed to be, right?

If yes:

Solution: sin( -cos( += 2 4

sinα-(cosα)=√2/4

sinα+cosα=√2/4

sin(π+cos(2π-α

sinα-cosα

sinα+cosα)

Solution: Since sin2a = 2 3, then cos2a = the root number of 3 5, because cos2a = the square of cosa - the square of sina = the square of 2 cosa - 1, so the square of cosa = (cos2a 1) 2 = 6 (root number 5 3), so sina = the root number of 6 (14 6 root number 5), so the square of cos(a faction 4) = (cosacos faction 4 sinasin faction 4) square = 1 2 (cosa sina) of the square.

cos²(α/4)？

1+cos(2a+π/2)/2

1-sin2a)/2

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∵3sin2a=2cosa

The square of both sides gets: 9sin 2a = 4cos a, that is, 9 (1-cos 2a) = 2 [2cos a-1]+2 9-9cos 2a = 2cos2a + 2

That is, 9cos 2a + 2cos2a-7 = 0

9cos2a-7)(cos2a+1)=0 solution: cos2a=7 9 or -1

Because 2 < <

So < 2

So cos2a = 7 9

i.e. 2cos a-1 = 7 9

So cosa = -2 times the root number 2 3

So cos( -=-cosa=2 times the root number 2 3

Because (0, 2), 2,0) so ( -0, )

Then sin( -0

sin(α-=4/5

Because ( 2,0).

So cos >0

cosβ=12/13

sinα=sin(α-

sin(α-cosβ+cos(α-sinβ=4/5×12/13-3/5×5/13

From (0, 2), 2,0),cos( -=3 5 can be judged ( -0, 2), then sin( -=4 5;

From ( 2,0), we get cos =12 13;

Finally, cos( -=3 5,sin( -=4 5,synchronously solved.

sinα=33/65