Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$

Question

Prove that

$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$

I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get it.

Answer 1

Hint: $\,\sin 30^\circ \sin 80^\circ - \sin 40^\circ \sin 50^\circ = \frac{1}{2} \sin 80^\circ - \sin 40^\circ \cos 40^\circ = \ldots\,$

Answer 2

Hint: $\sin(80^\circ)=\sin(2\cdot(40^\circ))=2\sin(40^\circ)\cos(40^\circ)$ and $\sin(30^\circ)=\frac12$.

Answer 3

Let us first prove that $\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$.

\begin{align} LHS & = \sin(30^\circ) \cdot \sin(80^\circ) \\ &= \cos(60^\circ) \cdot \cos(10^\circ) \\ & = \frac{1}{2} \cos(10^\circ) \\ \end{align}

\begin{align} RHS & = \sin(40^\circ) \cdot\sin(50^\circ) \\ & = \frac{1}{2}[\cos(40^\circ-50^\circ)-\cos(40^\circ+50^\circ)] \\ & = \frac{1}{2}[\cos(10^\circ) - \cos(90^\circ)] \\ & = \frac{1}{2} \cos(10^\circ) \\ \therefore LHS & = RHS \end{align}

So we have $$\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$$

And hence it follows that $$\frac{\sin(30^\circ)}{\sin(50^\circ)} = \frac{\sin(40^\circ)}{\sin(80^\circ)} $$

Answer 4

Let us start with a bit simplification by cross multiplying and replacing $\sin(30) =0.5$.

Then we should prove the following:

$$\frac{1}{2} \sin(80) = \sin(40)\sin(50)$$

Using $\sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)$, we can rewrite the right-hand side (Note: $\cos(x)=\cos(-x)$ since $\cos(x)$ is even function):

$$\frac{1}{2} \sin(80) = \frac{1}{2}\left(\cos(10)-\cos(90)\right),$$

where $\cos(90)=0$.

Now the question is if:

$$\sin(80) = \cos(10)$$

Which can be easily shown by:

$$\sin(80) = \sin(90-10) = \sin(90)\cos(10)-\cos(90)\sin(10)=\cos(10).$$

Note that I used the following to simplify the equation above:

$$ \sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$$

and the facts that $\sin(90)=1$ and $\cos(90)=0.$

Answer 5

$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$

$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{\sin (2(\cdot40^\circ))}$

$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{2 \cdot \sin 40^\circ \cdot \cos 40^\circ }$

$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{1}{2\cdot \cos 40^\circ }$

$\cos 40^\circ = \sin 50^\circ $

and that's true because $\sin x = \cos(\frac{\pi}{2}-x)$.