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I encountered this calculation in a problem $\dfrac{\sin 150^o\times\sin 20^o}{\sin 80^o\times\sin 10^o}$ and calculated that it equals 1.

Is it just a coincidence or is there any identity that says $\sin 150^o\times\sin 20^o=\sin 80^o\times\sin 10^o$?

I am trying to use the addition formulae and
$\sin \phi\sin \theta\equiv\dfrac{\cos (\phi-\theta)-\cos (\phi+\theta)}{2}$,
which reduces to showing $\cos 130^0\cos 170^0\equiv\cos 70^0\cos 90^0$, but still unable to explain why.

Any help is really appreciated. Many thanks!

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$\sin 150^0=\sin 30^0=1/2$

Hence LHS=$\frac{\sin 20^0}{2}$

$(\sin 2\theta)/2= \sin \theta \cos \theta\dots(1)$

$\sin 80^0=\cos 10^0 \implies$ RHS= $\sin 10^0 \cos 10^0$ = LHS from $(1)$

Hope this helps!

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We will need to utilize the following observations (the unit circle would help you see these visually):

  • $\sin 150^\circ=\frac 12$
  • $\cos 10^\circ = \sin 80^\circ$

Also the double-angle identity is used:

  • $\sin 20^\circ = 2 \cos 10^\circ \sin 10^\circ$ (double angle identity)

We apply these observations to prove your relation, from the LHS to the RHS. \begin{align} \sin 150^\circ \sin 20^\circ = \left(\frac 12\right) (2 \sin 10^\circ \cos 10^\circ) = \sin 10^\circ \cos 10^\circ=\sin 10^\circ \sin 80^\circ \end{align}

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You should know that $\sin 150^\circ=\sin 30^\circ=\frac 12$.

Now, $\sin 80^\circ=\cos 10^\circ$, because of the property $\sin x =\cos(90-x)$.

Now we have: $$\frac 12 \sin 20^\circ=\cos 10^\circ \times \sin 10^\circ$$ Recall the double angle identity $\sin(2x)=2\sin x\cos x$. This means that $\frac{\sin(2x)}{2}=\sin x\cos x$. Let's rewrite our equality: $$\frac{\sin 2(10)^\circ}{2}=\sin 10^\circ \times \cos 10^\circ$$ We can see that this is a double angle identity, where $x=10^\circ$. The LHS does indeed equal the RHS.

That is why $\sin 150^\circ \times \sin 20^\circ=\sin 80^\circ\times\sin 10^\circ$.

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