cosa-sina=0

or

cosa+sina=1

I was thinking you could square the bottom equation to get:

(cosa)^2 + 2cosasina + (sina)^2 = 1

Group terms to get:

(cosa)^2 + (sina)^2 + 2cosasina = 1

The first two terms form the identity:

(cosa)^2 + (sina)^2 = 1

So, using substitution:

1 + 2cosasina = 1

which reduces to:

2cosasina = 0

Now, the above forms the double angle identity:

sin2a = 2sinacosa

So, using substitution:

sin2a = 0

Now, we have to figure what angles would cause sin2a = 0:

0, +/- Pi/2, +/- Pi, +/- 3Pi/2,...

Normally, these angles would cause sin to equal 1 or -1, but since it is 2 times the angle it causes sin2a to equal zero.

a is written of the form:

a = +/- (n*Pi/2) for some n = 0, +/-1, +/-2,...

Part 2.

Do the same for the first equation:

(cosa)^2 - 2cosasina + (sina)^2 = 0

Again, use the identity:

(cosa)^2 + (sina)^2 = 1

Substitute:

1-2cosasina = 0

Add 2cosasina to both sides:

2cosasina = 1

Use the double angle formula:

sin2a =1

Now, we have to find what values will cause sin2a to equal 1:

Pi/4, 5Pi/4, 9Pi/4...

This is true because when the angle is multiplied by 2, it will be of the form n*Pi/2, which causes sin to equal 1 only when it is on the positive y axis. Otherwise, if 3Pi/4 were included, for example, it would cause sin2a to equal -1 because it would be the negative y axis.

So, a is expressed as:

a = n*Pi/4 for some n = 1,5,9,...