Known sin root number 2cos tan root number 3cot to find known belongs to 0, pie .

Since sin = root number 2cos, so sin 2 = 2cos 2, then cos 2 = 1-2cos 2

Because tan = root number 3cot

So tan 2 = 3 cot 2 = 3 cos 2 sin 2 = 3 cos 2 (1-cos 2).

And because tan 2=sin 2 cos 2 = 2cos 2 (1-2cos 2).

So 3cos2 (1-cos2) = 2cos2 (1-2cos2).

If cos = 0, then sin = 0, but , belong to (0, pie), so there is no solution at this time.

When cos is not 0, there will be 3-6cos 2=2-2cos 2

cosβ^2=1/4

cos = 1 2 or -1 2

But cos = -1 2.

sin = - root number 2 2<0, can belong to (0, pie), sin >0

So cos is not -1 2

cos can only be 1 2

60 degrees. Because tan = root number 3cot = 1, and belongs to (0, pie).

So =45 degrees.

In summary, =45 degrees, =60 degrees.

Because sin = root number 2< = 1

So cos <=1 2

Because tan = root number 3cot, therefore, belongs to (0, 1 2 * pie], so tan = root number [sin 2 (1-sin 2)], i.e. 3cot = 2cos (1-2cos) so 3cos sin = 2cos (1-2cos), i.e. cos = 0 or 2sin = 3 (1-2cos) The solution is = 90 degrees or =arccos(9 + root number 31 20) (> 1 2, round).

or =arccos[(9-root, number, 31) 20], so =0 degrees(round), or =arcsinroot[(9-root, number, 31) 10], in summary, =arcsinroot[(9-root: number: 31) 10], =arccos[(9-root: number: 31) 20].

Because sin = 2cos

So sin cos = 2, sin = 2cos because tan = 3ctg

Because sin = 2cos

So (sin) 2=2(cos) 2

Because cos = sin 6 3

So (cos) 2=2 3(sin) 2 add up: 2(cos) 2 2 3(sin) 2 16(cos) 2 2(sin) 2 34(cos) 2 1

cos = 1 2 or cos = -1 2cos =1 2, sin = 3 2, ctg = 3 3 then sin = 2 2, tan =1

cos = -1 2, sin = 3 2, ctg =- 3 3 then sin =- 2 2, tan =-1 in the fourth quadrant, rounded.

In summary, =45, =60

First of all, the first argument potato examines a formula sina+cosa = root number 2 and eggplant 3 about the same time squared, to get 1 + 2sinac0sa=2 3, near and hand excitation sinacosa = minus 1 6, sina-c0sa squared to get 1-2 sinacosa into the number of dramas The above formula is 4 3, and because a (0, 丌) is from the original formula 2 3 3 or -2 3 3

Both sides of the square have (sina) 2+6sinacosa+9(cosa) 2=10=10(sina) 2+10(cosa) 2, and the two tribes are cautious and divide by (cosa) 2.

tana) 2+6tana+9=10(tana) 2+10, solve tana=1 filial piety code 3

The square of both sides (sin -cos) = 2

That is, sin -2 sin cos +cos = 2 and sin +cos = 1

So -2sin cos = 1

i.e. sin2 = -1

and (0, )

So 2 (0,2).

Therefore 2 = 3 2

So =3 4

then tan = tan (3 4) = -1

sin -cos = root number 2 both sides squared at the same time yield: sina 2 + cosa 2-2sin *cos =2 i.e.: 1-sin2a=2 The answer is -1

sin -cos = root number 2 both sides squared at the same time yield: sina 2 + cosa 2-2sin *cos =2 i.e.: 1-sin2a=2 The answer is -1

Solution sina-cosa = 2

Both sides are squared:

sina-cosa)²=2

i.e., sin a-2sinacosa+cos a=2, i.e., 1-2sinacosa=2

2sinacosa=-1

sin2a=2sinacosa=-1

sin = 2cos, formula; tan = sin cos = 3cot = 3cos sin, two formulas;

1 2 2 formula: cos = 6sin 3, three formulas;

Combining one formula and three formulas obtains: sin +cos = ( 2cos ) 6sin 3) =2cos +2sin 3

2（cos²β+sin²β）4sin²β/3=2-4sin²β/3=1；4sin²β/3=1，sinβ=√3/2，cosβ=1/2；

sinβ=√2/2

From sin = root number 2cos, tan = root number 3cot can be known as cos = 2 3

Because 0< 0

So sin = 5 9

It can be seen that cos = 5 18

0< 0 again

So sin = 13 18