2. We are going to prove that E: 1; = \"flungllcos for a specic case. Let's assume ml; 6 R2 and .. . . . . . _. a . a hes along the pos1t1ve x-axrs. That Is to say, a = [0] for some arb1trary a > 0. Let's _, +b assign components to I; as b = [E] where bpbz 2 0 (but not both 0). The sign of the 2 components will tell you which quadrant the vector 1; lies in. For each quadrant, prove that 5-5 = "5"\"5" cos 0 (and if you do it for all four quadrants, well, then you did it for all of R2 ). Remember, 6 is the angle between (7 and 1;. The following trig facts will be helpful: adjacent 0 cos (9 = in a right triangle. hypotenuse 0 cos(l 80 t9) = cos(t9) 0 cos(t9) = cos(t9)