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Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.

My attempt that is not simple is as follows.

Expand both known constraints, so we have

\begin{align} \cos x \cos y &=4/15\\ \sin x \sin y &=-1/15 \end{align}

Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have

$$ 225 \sin^4 y -210 \sin^2 y +1=0 $$

with its solution $\sin^2 y = \frac{7\pm4\sqrt3}{15}$.

Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.

\begin{align} \sin^2(2y) &= 4\cos^2 y\sin^2 y\\ &= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\ \sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)} \end{align}

$\sin 2y$ must be negative.

Edit

Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.

By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is.

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  • $\begingroup$ Do you assume x and y are obtuse, or it is specified in the problem? $\endgroup$ – Quanto Sep 6 '19 at 15:24
  • $\begingroup$ @Quanto: Given. $\endgroup$ – Artificial Stupidity Sep 6 '19 at 15:25
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As $90<x,y<180$ and $\cos(x+y)>0$

$$270<x+y<360\implies\sin(x+y)=-\sqrt{1-(1/3)^2}$$

Again, $-90<x-y<90^\circ\implies\sin(x-y)=\pm\sqrt{1-(1/5)^2}$

Finally $$\sin2y=\sin(x+y+(x-y))=?$$

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Try to solve the first equation for $x$ and plug this in the second equation . I got this for $y$: $$\cos ^{-1}\left(\frac{1}{3}\right)=2 y+\cos ^{-1}\left(\frac{1}{5}\right)$$

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You might note that $7+4\sqrt3=(2+\sqrt3)^2$ and $2(2+\sqrt3)=(1+\sqrt3)^2$, and that your $\cos^2y$ has opposite sign to $\sin^2y$.

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$90<x,y<180^\circ$

$180<x+y<360\implies x+y=360-\arccos(1/3)$

If $x-y>0,x-y=\arccos(1/5)$

$\sin2y=\sin(360-\arccos(1/3)-\arccos(1/5))=-\sin(\arccos(1/3)+\arccos(1/5))$

Now $\arccos(1/3)+\arccos(1/5)=\arcsin(2\sqrt2/3)+\arccos(2\sqrt6/5)$

Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

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