The sines speak otherwise

$\begin{cases} \dfrac {\sin x}{\sin y} = 3 & \implies \sin x = 3 \sin y & \implies \sin^2 x = 9 \sin^2 y \\ \dfrac {\cos x}{\cos y} = \dfrac 12 & \implies \cos x = \dfrac {\cos y}2 & \implies \cos^2 x = \dfrac {\cos^2 y}4 \end{cases}$

$\begin{aligned} \implies 9 \sin^2 y + \frac {\cos^2 y}4 & = \sin^2 x + \cos^2 x \\ 9 \sin^2 y + \frac {\cos^2 y}4 & = 1 \\ 36 \sin^2 y + \cos^2 y & = 4 \\ 35 \sin^2 y + 1 & = 4 \\ \sin^2 y & = \frac 3{35} \end{aligned}$

Since $\cos 2y = 1 - 2\sin^2 y = 1 - 2 \times \dfrac 3{35} = \dfrac {29}{35}$ .

Note that $\sin^2 x = 9 \sin^2 y = \dfrac {27}{35}$ $\implies \cos 2x = 1 - 2 \times \dfrac {27}{35} = - \dfrac {19}{35}$

Therefore, we have:

$\begin{aligned} \frac {\sin 2x}{\sin 2y} - \frac {\cos 2x}{\cos 2y} & = \frac {2\sin x \cos x}{2 \sin y \cos y} - \frac {-\frac {19}{35}}{\frac {29}{35}} \\ & = 3 \times \frac 12 + \frac {19}{29} \\ & = \frac {125}{58} \end{aligned}$

$\implies p + q = 125 + 58 = \boxed{183}$

Write sin x = 3k, sin y=k, cosx = (1-9k^2)^0.5 = l, and cos y = (1-k^2)^0.5=2l, where l, and m are constants.Solve the simultaneous eqs. l^2 = 1-9k^2, and 4l^2 = 1-k^2, then we find k^2 = 3/35.

(sin 2x/sin 2y) - (cos 2x/cos 2y) = (2sin x.cos x)/(2sin y . cos y) - (2 (cos x)^2 - 1) /(2(cos y)^2 - 1) = 3/2 + 19/29 = 125/58=p/q, so p+q = 183.

you gonna use cos(2x)= 1 − 2 sin^2(x) and cos(2y)= 1 − 2 sin^2(y) and sin(y)= 1/3 sin (x) cos(y)= 2 cos(x) sin(y)^2 + cos(y)^2 -> sin(x) sin(x)^2 + cos(x)^2 -> sin(y) 3/2 + 19/29 = 125/58= p/q then p+q = 183 Done!

we know that sin^2(x)+cos^2(x)=1 let it be eqn 1 and from problem sin(x)=sin(y)×3 squaring on both sides we get sin^2(x)=9sin^2(y) and similatly other equn in prob is cos (x)=cos(y)÷2 and squqring on bth sides we get cos^2(x)=cos^2(y)÷4 substituting sin^2(x) and cos^2(x) values in eqn1 we get eqn in tetms of sin^2(y) and cos^2(y) let this be eqn 2 and we we know that sin^2(y)+cos^2(y)=1 let it be eqn 3 solving eqn 2 and 3 we get sin^2(y) and cos^2(y) values and from these we can get sin^2(x) and cos^2(x) values substituting we know that sin2x=2sinx×siny and cos2x=2cos^2(x)-1 substituting all values on req exp we will get its value