#sin^2a+cos^2a=#?
1 Answer
Jun 21, 2018
Explanation:
#"consider the right triangle with sides x and y and "#
#"hypotenuse r, a is the angle between x and r"#
#sina=y/r" and "cosa=x/r#
#r^2=x^2+y^2larrcolor(blue)"Pythagoras' Identity"#
#sin^2a+cos^2a=(y/r)^2+(x/r)^2#
#color(white)(xxxxxxxxxx)=y^2/r^2+x^2/r^2#
#color(white)(xxxxxxxxxx)=(x^2+y^2)/r^2=(x^2+y^2)/(x^2+y^2)=1#
