Show $\frac{\sin20^\circ\sin30^\circ\tan40^\circ}{\sin20^\circ\sin30^\circ+\tan40^\circ\sin50^\circ} =\tan10^\circ$

Question

The question arose in finding $\alpha$ in this diagram:

_{(source: mei.org.uk)}

With a bit of work with the sine rule it can be shown that:

$$\tan(\alpha) = \frac{\sin(20^\circ)\sin(30^\circ)\tan(40^\circ)}{\sin(20^\circ)\sin(30^\circ)+\tan(40^\circ)\sin(50^\circ)}$$

... and this can be evaluated on a calculator to find $\alpha = 10^\circ$. But I'm not satisfied in using a calculator.

I want to know if there's a neat way of manipulating the expression on the RHS so that it simplifies out to $\tan(10^\circ)$.

I've had a good go at it - using double/triple/quadruple angle formulae to re-write everything in terms of $\sin(10^\circ)$, $\cos(10^\circ)$ and $\tan(10^\circ)$ and trying to simplify - but just seem to make things more complicated.

Can you find a nice way of doing it?

Answer 1

Divide the numerator and denominator by $\sin20\sin30= \frac12\sin20$

\begin{align} &\frac{\sin20\sin30\tan40}{\sin20\sin30+\tan40\sin50}\\ =& \frac{\sin40}{\cos40+4\sin50\cos20} = \frac{\sin40}{(\cos40+1)+2\cos20} =\frac{\sin20}{\cos20+1}= \tan10\\ \end{align} where the identities $\sin2t=2\sin t\cos t$, $\cos2t+1=2\cos^2t$ are used twice each.

Answer 2

Edit: A simpler method. Thanks to @Blue, for their comment.

$\begin{align}\frac{\sin20^\circ\sin30^\circ\tan40^\circ}{\sin20^\circ\sin30^\circ+\tan40^\circ\sin50^\circ} &= \frac{\sin20^\circ\sin30^\circ\tan40^\circ}{\sin20^\circ\sin30^\circ+\cot50^\circ\sin50^\circ} \\ &=\tan40^\circ\frac{\sin20^\circ\sin30^\circ}{\sin20^\circ\sin30^\circ+\cos50^\circ} \\ &=\tan40^\circ\frac{\sin20^\circ\sin30^\circ}{\sin20^\circ\sin30^\circ+\cos(20^\circ+30^\circ) } \\ &=\tan40^\circ\frac{\sin20^\circ\sin30^\circ}{\sin20^\circ\sin30^\circ+(\cos20^\circ\cos30^\circ-\sin20^\circ\sin30^\circ) } \\ &=\tan40^\circ\frac{\sin20^\circ\sin30^\circ}{\cos20^\circ\cos30^\circ} = \tan20^\circ\tan30^\circ\tan40^\circ\\ & = \tan10^\circ\end{align}$

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Now for $\tan20^\circ\tan30^\circ\tan40^\circ = \tan10^\circ$:

Using, $\tan \alpha\tan(60^\circ-\alpha)\tan(60^\circ+\alpha) = \tan(3\alpha)$

Taking
$\begin{align}&\alpha = 10^\circ \\& \tan10^\circ\tan50^\circ\tan70^\circ = \tan30^\circ \\\Rightarrow& \tan10^\circ\dfrac{1}{\tan40^\circ\tan20^\circ} = \tan30^\circ \\\Rightarrow&\tan20^\circ\tan30^\circ\tan40^\circ = \tan10^\circ\end{align}$

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Earlier method:

$\small{\begin{align}\tan\alpha &= \frac{\sin20^\circ\sin30^\circ\sin40^\circ}{\sin20^\circ\sin30^\circ\cos40^\circ+\sin40^\circ\sin50^\circ} \\\\ &= \frac{\sin20^\circ\sin30^\circ\sin40^\circ}{\sin20^\circ\sin30^\circ\sin50^\circ+\sin40^\circ\sin50^\circ}\\\\ &= \frac{\sin20^\circ\sin30^\circ\sin40^\circ}{\sin50^\circ(\sin20^\circ\sin30^\circ+\sin90^\circ\sin40^\circ)} & (\sin90^\circ = 1)\\\\ &= \frac{2\sin20^\circ\sin30^\circ\sin40^\circ}{\sin50^\circ((\cos10^\circ-\cos50^\circ)+(\cos50^\circ-\cos130^\circ))}&(\small{2\sin x\sin y = \cos(x-y)-\cos(x+y)})\\\\ &=\frac{2\sin20^\circ\sin30^\circ\sin40^\circ} {\sin50^\circ(2\sin60^\circ\sin70^\circ)}&(\cos x-\cos y =2\sin\frac{x+y}2\sin\frac{x-y}2 ) \\\\ &=\frac{\sin20^\circ\sin30^\circ\sin40^\circ}{\cos20^\circ\cos30^\circ\cos40^\circ} =\tan20^\circ\tan30^\circ\tan40^\circ \\\\ &= \tan10^\circ\end{align}}$

Answer 3

Just for the fun of it, consider the equation $$\frac{\sin (2 x) \sin (3 x) \tan (4 x)}{\sin (2 x) \sin (3 x)+\sin (5 x) \tan (4 x)}=\tan(x)$$ Let $x=\tan^{-1}(t)$ and this write $$\frac{4 (t-1) (t+1) \left(t^2-3\right) t}{\left(3 t^2-1\right) \left(t^4-10 t^2+13\right)}+t=0$$ Discarding the trivial $t=0$, this reduces to $$\frac{3 t^6-27 t^4+33 t^2-1}{\left(3 t^2-1\right) \left(t^4-10 t^2+13\right)}=0$$

The numerator is a cubic in $t^2$ with three real roots since $\Delta=331776$. Using the trigonometric method (!!), they are $$t^2_k=3+\frac{8}{\sqrt{3}} \cos \left(\frac{\pi }{18}-\frac{2 \pi k}{3}\right) \qquad \text{for} \qquad k=0,1,2$$ They are all positive and the smallest (corresponding to $k=2$) is $$t^2_2=3-\frac{8}{\sqrt{3}}\sin \left(\frac{2 \pi }{9}\right)\implies x=\tan^{-1}\Bigg[\sqrt{3-\frac{8}{\sqrt{3}}\sin \left(\frac{2 \pi }{9}\right) } \Bigg]$$

At this point, I am totally stuck but numerically this is $\frac \pi {18}$