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$$ \begin{aligned} \sin x=2\cdot\sin 10^\circ \sin (20^\circ+x) \Rightarrow \tan x&=\dfrac{2\cdot\sin 10^\circ\sin 20^\circ}{1-2\cdot\sin 10^\circ \cos 20^\circ}\\&=\dfrac{2\cdot\sin 10^\circ \sin 20^\circ}{2\cdot\cos 10^\circ \sin 20^\circ}\\&=\tan 10^\circ \end{aligned} $$

I saw this on https://math.stackexchange.com/a/516019/1228919, but I don't understand how to prove this.

I know that: $$ \sin(\alpha+\beta)=\sin \alpha\cdot\cos \beta+\cos \alpha\cdot\sin \beta \\[6pt] \text{ and }\\[6pt] \tan \alpha =\dfrac{\sin \alpha}{\sin \beta} $$

But finally I got this: $$ \tan x=\dfrac{2\sin 10^\cdot\sin 20^\circ\cos x+2\sin 10^\circ\sin x\cos 20^\circ}{\cos x} $$

Guilherme Costa
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Nonus
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  • Do you know how to expand $\sin (A+B)$ via the addition formula? – Calvin Lin Oct 03 '23 at 02:26
  • The equation is wrong, it should be $\sin(x) = 2\sin(10º)\sin(20º+ x)$. – Trobeli Oct 03 '23 at 02:27
  • Hint: Expand using the formula $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$, then divide by $\cos(x)$ in both sides, and solve for tan(x) to get the desired expression. Finally, to transform that expression, use Prosthaphaeresis formula's (or sum-to-product formula) conveniently to get the result. – Trobeli Oct 03 '23 at 02:32

1 Answers1

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We are going to use the following items:

  1. $\tan(A) = \frac{\sin(A)}{\cos(A)}$,
  2. $\sin(A+B) = \sin(A)\cos(B) + \sin(B)\cos(A)$,
  3. $\sin(30^{\circ}) = \frac{1}{2}$.

Diving both sides of first equation by $\sin(x)$, we have $$ 1 = 2\frac{\sin(10^{\circ})(\sin(20^{\circ})\cos(x) + \sin(x)\cos(20^{\circ}))}{\sin(x)} =2\sin(10^{\circ})\left(\frac{\sin(20^\circ)}{\tan(x)} + \cos(20^\circ)\right). $$

Solving for $\tan(x)$, we have that

$$ \begin{align*} 1 &= 2\sin(10^\circ)\cos(20^\circ) + 2\frac{\sin(10^\circ) \sin(20^\circ)}{\tan(x)}\\ \Rightarrow& 1 - 2\sin(10^\circ)\cos(20^\circ) = 2\frac{\sin(10^\circ) \sin(20^\circ)}{\tan(x)}\\ \Rightarrow& \tan(x) = 2\frac{\sin(10^\circ) \sin(20^\circ)}{1 - 2\sin(10^\circ)\cos(20^\circ)}. \end{align*} $$

Now, remember that $$\frac{1}{2} = \sin(30^\circ) = \sin(10^\circ + 20^\circ) = \sin(10^\circ)\cos(20^\circ) + \sin(20^\circ)\cos(10^\circ).$$

Using this, we have

$$\tan(x) = 2\frac{\sin(10^\circ) \sin(20^\circ)}{1 - 2\sin(10^\circ)\cos(20^\circ)} = \frac{2\sin(10^\circ) \sin(20^\circ)}{2\sin(20^\circ)\cos(10^\circ)} = \tan(10^\circ).$$

Guilherme Costa
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