# Sin x / sin y = 3 and cos x / cos y = 1/2 find the value of sin 2x/sin 2y + cos 2x/cos 2y

Vikas TU
14149 Points
6 years ago
Write siny = sinx/3  and squaring both sides,
(siny)^2 = (sinx)^2/9..............(1)
(cosy)^2 = 4(cosx)^2........(2)
and equate it to 1.
and solve in sinx or cosx in any of them by writing cosx = 1 – sin^2x or vice versa.
get x values and then simply y values. and then u can find the final value.

Abhinav Anand
10 Points
6 years ago
Ans is 49/58 and I am not getting. Please be little more specific. Thanks in Advance And if possible show working .
govardhan
18 Points
6 years ago
siny=sinx/3
sin^2y=sin^2x/9
similarly,
cos^y=4cos^2x
then,sin^2y+cos^2y=sin^2x/9+4cos^2x
1=(sin^2x+36cos^2x)/9
9=sin^2x+36cos^2x
sin^2x+36(1-sin^2x)=9
by solving
sinx=27/35
now,sin^2x+cos^2x=36sin^2y+cos^2y
4=36(1-cos^2y)+cos^2y
by solving cosy=32/35

now, given that
sinx/siny=3 and cosx/cosy=1/2
sinx=3siny-------(1) and 2cosx=cosy---------(2)
multiplying (1) and (2)
(sinx)(2cosx)=3sinycosy
sin2x=(3/2)sin2y
sin2x/sin2y=3/2
and
sin^2x=27/35
cos^2x=1-27/35=8/35
cos^2y=32/35
sin^2y=1-32/35=3/35

and finally,
(sin2x/sin2y)+(cos2x/cos2y)=(3/2)+[(8/35)-(27/35)]/[(32/35)-(3/25)]
=3/2-19/29
=49/58

Srk
12 Points
5 years ago
siny=sinx/3sin^2y=sin^2x/9similarly,cos^y=4cos^2xthen,sin^2y+cos^2y=sin^2x/9+4cos^2x1=(sin^2x+36cos^2x)/99=sin^2x+36cos^2xsin^2x+36(1-sin^2x)=9by solvingsinx=27/35 now,sin^2x+cos^2x=36sin^2y+cos^2y 4=36(1-cos^2y)+cos^2yby solving cosy=32/35 now, given thatsinx/siny=3 and cosx/cosy=1/2sinx=3siny-------(1) and 2cosx=cosy---------(2)multiplying (1) and (2)(sinx)(2cosx)=3sinycosysin2x=(3/2)sin2ysin2x/sin2y=3/2 andsin^2x=27/35cos^2x=1-27/35=8/35cos^2y=32/35sin^2y=1-32/35=3/35 and finally,(sin2x/sin2y)+(cos2x/cos2y)=(3/2)+[(8/35)-(27/35)]/[(32/35)-(3/25)] =3/2-19/29 =49/58