In a right triangle, relative to a $45^\circ$ angle, if we have
$$\text{adjacent} = 1$$
$$\text{opposite} = 1$$
then
$$\text{hypotenuse}=\sqrt{o^2+a^2}=\sqrt{1^2+1^2}=\sqrt{2}$$
so that
$$\sin 45^\circ=\frac{1}{\sqrt{2}}$$
But, when
$$\text{hypotenuse} = 1$$
$$\text{opposite} = \text{adjacent}$$
then (writing $o$ for $\text{opposite}$ and $a$ for $\text{adjacent}$)
$$\begin{align}
o &= \sqrt{h^2-a^2} = \sqrt{1^2-o^2} \\[4pt]
\implies \quad o^2 &= h^2-o^2\\
&=1-o^2 \\[4pt]
\implies\quad 2o^2&=1 \\
\implies\quad o&=\sqrt{\frac{1}{2}}
\end{align}$$
so that
$$\sin 45^\circ =\frac{\sqrt{\frac{1}{2}}}{1}$$
What went wrong?
Best Answer
Nothing went wrong. $$ \frac{\sqrt{\frac{1}{2}}}{1} = \sqrt{\frac12}=\frac1{\sqrt2} $$