If a right $\triangle ABC$ has $\angle A= 90^\circ$, $\angle B=45^\circ$, $\angle C=45^\circ$. Is there a way of finding the lengths of the sides $a$, $b$ & $c$ without knowing any of their lengths ? Normally we use $(\cos{x})^2+(\sin{x})^2=1$ as the hypotenuse.
I have an equation that says: $$(\sin {x}\times \sin {x}\times \cos {x})^2+(\cos {x}\times \cos {x}\times \sin {x})^2=(\sin {x}\times \cos {x})^2$$
$$\frac{(\sin {x}\times \sin {x}\times \cos {x})^2}{(\sin {x}\times \cos {x})^2}=(\sin{x})^2$$
$$\frac{(\cos {x}\times \cos {x}\times \sin {x})^2}{(\sin {x}\times \cos {x})^2}=(\cos{x})^2$$
Best Answer
Here's an interesting fact:
Specifically, the "$\sin A$-$\sin B$-$\sin C$" triangle is the one inscribed in a circle of diameter $1$; you might consider it a fundamental representative of the family of triangles with angles $A$, $B$, $C$, but it is not the only triangle with those angles. You can get from the representative to any other member of the family by magnifying the side-lengths (and the circumdiameter) by some factor, $m$; conversely, any member of the family has side-lengths that are multiples of the side-lengths of the representative (since the family members are all similar).
This is, in fact, exactly what the Law of Sines tells you: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = m$$ where $m$ is that scale factor ... and also the diameter of the triangle's circumcircle.
In the case of a right triangle with hypotenuse $1$, recall Thales' Theorem:
Thus, an hypotenuse-$1$ right triangle is inscribed in a circle whose diameter matches that hypotenuse: in other words,
Such a triangle's sides are indeed $\sin A$, $\sin B$ (which we also call "$\cos A$", since $A$ and $B$ are complementary), and $\sin 90^\circ$ (which, of course, is $1$). Perhaps this is what you were getting-at in your comment to your question.