How to find the side of the square ,by using trigonometry There is a square with diagonal length of 'a", the question is to find the length of the sides. it can be found by Pythagorean theorem. but I tried to do it with trigonometry, considering the properties of a square but it doesn't resemble the first answer, so I was wondering what did I forget/mistake?
 A: Hint:
A side is an orthogonal projection of the diagonal, onto a line making  an angle of $\pi/4$ with the diagonal.
A: Let the side length of the square equal $b$, and the hypotenuse $a$. By Pythagoras' Theorem, we have
\begin{align}
b^2+b^2&=a^2 \\
2b^2&=a^2 \\
b^2&=\frac{a^2}{2}\\
b&=\sqrt{\frac{a^2}{2}}=\frac{\sqrt{a^2}}{\sqrt 2}=\frac{a}{\sqrt 2}=\frac{\sqrt 2}{2}a
\end{align}
Alternatively, the angle between $a$ and $b$ is $45$ degrees. We know this because a square has $4$ right angles, which are bisected (cut in half) by the diagonal. Therefore,
\begin{align}
\sin(45)&=\frac{b}{a}\\
\end{align}
Now find the value of $\sin(45)$, and rearrange the equation to find $b$. Hopefully this answer is familiar.
It seems that part of your confusion stems from the fact that you are unsure about rationalising the denominator:
$$
\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}
$$
Consider the left-hand side of the above equality. Multiplying $\frac{1}{\sqrt 2}$ by $1$ does not change its value:
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}
$$
We also know that
$$
1=\frac{\sqrt{2}}{\sqrt{2}}
$$
Therefore,
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt 2}{2}
$$
A: The diagonal makes a $45^\circ=\frac{\pi}{4}$ angle with all vertices it touches with its endpoints.
If you look at trig identities here,  you will see that $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\approx 0.707$ no matter which corner or which side of the diagonal you chose. The length of each side is $\frac{\sqrt{2}a}{2}$.
