The unit circle is a circle with a radius of 1. Because of this, the coordinates of any point at this circle will correspond to (cosΘ, sinΘ). Where Θ is the given angle.
In this case, you have Θ=π/4
Then you need to find the values for sin(π/4) and cos(π/4) to find the point
First, let's convert π/4 into degrees

Now imagine a right triangle with one of the acute angles set to 45°, the internal angles of a right triangle have to sum 180°, then if you have a 90° and a 45° angle, then the other one will be 180°-90°-45°=45°. Then you will have an isosceles triangle.
By setting the length of one of the sides adjacent to the right angle to
1 and applying the Pythagorean theorem, you'll find the length of the hypotenuse as follows:
![\begin{gathered} a^2+b^2=h^2 \\ 1^2+1^2=h^2 \\ h=\sqrt[]{1^2+1^2}=\sqrt[]{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/588z60uqb6y9lg3p6j5aoje6w3nf1oqbui.png)
Then the cosine is given by:
![\begin{gathered} \cos (45)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{\sqrt[]{2}}\text{ multiply the numerator and denominator by }\sqrt[\square]{2} \\ \cos (45)=\frac{1\cdot\sqrt[]{2}}{\sqrt[]{2}\cdot\sqrt[]{2}}=\frac{\sqrt[]{2}}{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wjygmg4hoowsnd5a9zl67i7u2e0o2meods.png)
Now you can find the sine by:
![\begin{gathered} \sin (45)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1}{\sqrt[]{2}}\text{ multiply the numerator and denominator by }\sqrt[\square]{2} \\ \sin (45)=\frac{1\cdot\sqrt[]{2}}{\sqrt[]{2}\cdot\sqrt[]{2}}=\frac{\sqrt[]{2}}{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8izxzvnor24vt8ukjaqh9d5sscp9f16ulu.png)
Then the coordinates of the point will be (cos45, sin45) = (√2/2, √2/2)