Known sin sin 1 4, cos cos 1 3, find tan

sin + sin = 1 4 and the difference product.

sin +sin = 2sin[( 2]cos[( 2] 1 4cos +cos =1 3 and differential product.

cos +cos = 2cos[( 2]cos[( 2] 1 3 ( + 2=x

So divide the two formulas by tanx=3 4

Therefore, tan( +=tan2x=2tanx [1-(tanx) 2]=12 7

tan(α+= sin(α+/cos(α+sin(α+=sinα·cosβ+cosα·sinβ;

cos( +=cos ·cos -sin ·sin sin +sin =1 4 so (sin +sin ) 2 = 1 16;

i.e. sin 2 +sin 2 +2sin sin =1 16cos +cos =1 3 so (cos +cos) 2 = 1 9

That is, cos 2 +cos 2 +2cos cos = 1 9, let's think for yourself.

I think that's all there is to it, it's simple.

From sin +sin =1 4: 4sin +4sin =1 by cos +cos =1 3: 3cos +3cos =1 by , get:

4sin +4sin =3cos +3cos i.e.: 4sin -3cos =3cos -4sin i.e.: sin( -sin( -sin ( -where is the acute angle, sin =3 5, cos =4 potato quiet5 According to the formula, there are two cases:

In the first case, -2k + k=0, 1, 2....At this time, +2k +2 , so: tan( +tan2 =2tan (1-(tan ) 2) =24 7 In the second case, -2k-1) - at this time, =2k-1) + from this we get:

sin = -sin and cos = -cos, which is inconsistent with the known conditions and is discarded. Count the slag.

tanα=1/4,tan(α-tanα-tanβ)/1+tanαtanβ)=1/3

1 4-tan ) 1+tan 4) = 1 31-4tan ) partial deficiency (4+tan ) = 1 34 + tan = 3-12 tan

13tanβ=-1

tanβ=-1/13

I wish Chan Latan to learn to enter the He Tong step.

From sin +sin =1 4: 4sin +4sin =1 by cos +cos =1 3: 3cos +3cos =1 by , get:

4sin +4sin =3cos +3cos i.e.: 4sin -3cos =3cos -4sin i.e.: sin( -sin( -

The formula is an acute angle, sin = 3 5, cos = 4 potato 5 according to the formula, divided into two cases:

In the first case, -2k + k=0, 1, 2....

At this point, +2k +2 , so:

tan( +tan2 =2tan (1-(tan ) 2) second case, -2k-1) -

In this case, =2k-1) +

From this: sin = -sin and cos = -cos, which is inconsistent with the known conditions of the grasp, discarded. Count the slag.