IIT JEE , AIEEE Forum - Mathematics , Trignometry

sti

0<x<y<pie/4. sin(y-x)=5/13, cos(y+x)=4/5. tan 2y=?????????????????

master_purav

cos (y-x) = 12/13

sin (y+x) = 3/5

(Using cos^2 + sin^2 = 1)

Sin (2y) = Sin (y+x+y-x) = Sin (y+x)Cos(y-x) + Sin (y-x)Cos(y+x)

Cos (2y) = Cos (y+x+y-x) = Cos (y+x)Cos(y-x) - Sin (y+x)Sin (y-x)

Get Tan 2y = Sin / cos

Note: Here all angles are in first quadrants and so we get all sines and cosines +ve...

hsbhatt

Another Method:
y - x = sin-1(5/13) = tan-1(12/13)
y + x = cos-1(4/5) = tan -1(3/4)

giving 2y = tan-1(12/13) + tan -1(3/4)

Then using tan(A+B) formula, we get

Tan2y = [12/13+3/4]/[1-12/13*3/4] = -87/16

sprinkle

sin(y-x)=5/13 => cos(y-x) = 12/13 => tan(y-x) = 5/12

cos(y+x)=4/5 => sin(y+x) = 3/5 => tan(y+x) = 3/4

tan 2y = tan ((x+y) + (y-x)) = (tan(x+y) + tan(y-x))/(1 - tan(x+y)*tan(y-x))

                                        = (3/4 + 5/12)/(1 - (3/4) * (5/12))

                                        = (14/12)/(11/16) = 56/33

   (sorry for mis-calculations if any

, as ans is not matching with previous ones)

hsbhatt

I am sorry. Was a bit drowsy when posting this one. sprinkle's calculation's right

Ask & Discuss Questions with Community & Experts