Look at the picture.
Use the Pythagorean theorem:
[tex]b^2=a^2+a^2\\\\b^2=2a^2\to b=\sqrt{2a^2}\\\\b=\sqrt{a^2}\cdot\sqrt2\\\\b=a\sqrt2[/tex]
[tex] \sin\theta=\dfrac{opposite}{hypotenuse}\\\\\theta=45^o,\ opposite=a,\ hypotenuse=a\sqrt2 [/tex]
substitute
[tex]\sin45^o=\dfrac{a}{a\sqrt2}=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=\dfrac{\sqrt2}{2}[/tex]
Answer: [tex]\sin45^o=\dfrac{\sqrt2}{2}[/tex]
