It is known that Tan3 is found to be 1 4sin cos 3sin 5cos

Because tan = 3 so sin = 3cos, according to the Pythagorean theorem: (sina) 2+(cosa) 2=1, the simultaneous solution obtains: cosa = plus or minus (10 under the root number) 10, substituting the following equation can be obtained.

4sinα-cosα /3sinα +5cosα12cosα-1/9+5cosα

17cosα-1/9

Because tan = 3 so sin = 3cos 4sin -cos 3sin +5cos 4sin -1 9+5sin 3

17sinα/3-1/9

sin xsin +cos xcos = 1 so 10 sin xsin 9 = 1 so calculate sin in the substitution sub.

Note that there are two answers (plus or minus ha).

tanα=3,sinα=3cosα ,sin²α=9cos²α,sinα=±3√10/10,cosα=±10/10

sin +cos ) 1+2sin cos =1+6 10=8 5, or

sin +cos ) 2 5, or ,1,sin = 3 10 10, cos = 10 10, how are these two calculated, can you explain it, troublesome, known tan = 3, find (sin +cos).

The speed is in a hurry, and it is better to take some steps.

Equal to 0 ; Limb Zheng commanded.

3sin -tan )=3cos *tan -tan =3*3 of the plex stupid 1*tan -tan =tan -tan =0

The denominator is 0!The answer is 0!

Summary. 2cos +3sin 4sin -5cos up and down at the same time by cosa=(2tana+3) (4tana-5) and substituting tana=-1 3 into (-2 3+3) (4 3-5)=7 3 (-19 3) =7 3 (-3 19)=-7 19

Knowing that tan = -1 3, find the value of 2cos +3sin 4sin -5cos.

ok2cos +3sin 4sin -5cos up and down at the same time except cosa=(2tana+3) (4tana-5) shouting stool to replace tana=-1 3 into Dexiaohong (skillfully infiltrated book-2 3+3) (4 3-5)=7 3 (-19 3) =7 3 (-3 19)=-7 19

From tan = -1 3, sin = -1 (root number 10), cos = 3 (root number 10).

So it is easy to find: 7sin -3cos 4sin +5cos =8 (root number 10) + 9 4