Cosx + cosy = 1 / 2, SiNx siny = 1 / 3, then cos (x + y) =?

Cosx + cosy = 1 / 2, SiNx siny = 1 / 3, then cos (x + y) =?

With cos (x + y) = cosxcosy sinxsiny, add the left and right squares of the formula given by the topic to the left and right of the two formulas as follows: cos ^ 2x + cos ^ 2Y + 2cosxcsoy + sin ^ 2x + sin ^ 2y-2sinxsiny = 13 / 36. From cos ^ 2x + sin ^ 2x = 1, cos ^ 2Y + sin ^ 2Y = 1, 1 + 2cosxcsoy + 1-2sinxsiny

SiNx siny = - √ 3 / 3 cosx cosy = - 1 / 3 then cos (X-Y)

sinx-siny=-√3/3
Square of both sides at the same time:
sin ² x-2sinxsiny+sin ² y=1/3 ①
cosx-cosy=-1/3
Square of both sides at the same time:
cos ² x-2cosxcosy+cos ² y=1/9 ②
From ① + ②:
-2sinxsiny-2cosxcosy=4/9
That is - 2 (sinxsiny + cosxcosy) = 4 / 9
-2cos(x-y)=4/9
cos(x-y)=-2/9
Answer: - 2 / 9

Given SiNx siny = - 1 / 3, cosx cosy = 1 / 2, find cos (X-Y) =

sinx-siny=-1/3,（1） cosx-cosy=1/2 （2）(1) ²+ （2） ² sin ² x+sin ² y-2sinxsiny+cos ² x+cos ² y-2cosxcosy=1/9+1/42-2(cosxcosy+sinxsiny)=13/362cos(x-y)=2-13/36=59/36cos(x-y)=59/72...

Cos2x / radical 2cos (x + π / 4) = 1 / 5,0

(cos2x)/√2cos(x＋π/4)＝(cos ² x－sin ² x) / [√ 2 (cosxcos (π / 4) - sinxsin (π / 4))] = [(cosx + SiNx) (cosx SiNx)] / (cosx SiNx) = cosx + SiNx = 1 / 5

Master simplification f (x) = 2 * root 3 * SiNx * cosx + 2cosx square - 1

F (x) = 2 * root 3 * SiNx * cosx + 2cosx square - 1=
=√3sin2x+cos2x
=2*(√3/2sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)

How do you get f (x) = SiNx + cosx = radical 2Sin (x + Pai / 4)?

f(x) = sinx+cosx
= √2(sinx (1/√2) + cosx (1/√2) )
= √2(sinx cosπ/4 + cosx sinπ/4)
= √2(sin(x+π/4))