Answer :
Look at the attached figure. We define the unit circle as a circle with center [tex]A[/tex] in the origin (0,0) and radius 1.
Then, we consider a point P on the circumference. We call [tex]\alpha[/tex] the angle between the positive half of the x axis and the radius AP.
We define
[tex]\cos(\alpha) = \overline{AD},\quad \sin(\alpha) = \overline{AC}[/tex]
As you can see, ACD is a right triangle, and so we have
[tex]\overline{AD}^2+\overline{AC}^2=\overline{AP}^2[/tex]
But since we know that AD is the cosine, AC is the sine, and AP is the radius (which is 1, and remains 1 when squared), we have just found out that
[tex]\cos(\alpha)^2+\sin(\alpha)^2=1[/tex]
Answer: Here is a copy paste-able, Edg2021 friendly version of the above answer^ :)
I defined the unit circle as a circle with center A in the origin (0,0) and radius 1. Then, I considered a point P on the circumference. I call a the angle between the positive half of the x axis and the radius AP. I defined that
cos(a)= line AD, sin(a)= line AC
ACD is a right triangle, and so,
line AD^2+ line AC^2 = line AP^2
But since I knew that AD is the cosine, AC is the sine, and AP is the radius (which is 1, and remains 1 when squared), I found that
cos(a)^2 + sin(a)^2 = 1