# Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$?

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.

• Do you assume x and y are obtuse, or it is specified in the problem? – Quanto Sep 6 '19 at 15:24
• @Quanto: Given. – Artificial Stupidity Sep 6 '19 at 15:25

As $$90 and $$\cos(x+y)>0$$

$$270

Again, $$-90

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

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

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 $$\sin^2y$$.

$$90

$$180

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