Vtyjkyuk

Prove that

$$fracsin 30^circsin 50^circ= fracsin 40^circsin 80^circ$$

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.

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

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

beginalign
LHS & = sin(30^circ) cdot sin(80^circ) \
&= cos(60^circ) cdot cos(10^circ) \
& = frac12 cos(10^circ) \
endalign

beginalign
RHS & = sin(40^circ) cdotsin(50^circ) \
& = frac12[cos(40^circ-50^circ)-cos(40^circ+50^circ)] \
& = frac12[cos(10^circ) - cos(90^circ)] \
& = frac12 cos(10^circ) \
therefore LHS & = RHS
endalign

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

And hence it follows that $$fracsin(30^circ)sin(50^circ) = fracsin(40^circ)sin(80^circ) $$

Hint: $,sin 30^circ sin 80^circ - sin 40^circ sin 50^circ = frac12 sin 80^circ - sin 40^circ cos 40^circ = ldots,$

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

Then we should prove the following:

$$frac12 sin(80) = sin(40)sin(50)$$

Using $sin(a)sin(b)=frac12left(cos(a-b)-cos(a+b)right)$, we can rewrite the right-hand side (Note: $cos(x)=cos(-x)$ since $cos(x)$ is even function):

$$frac12 sin(80) = frac12left(cos(10)-cos(90)right),$$

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 $cos(90)=0.$