cosa-sina=0
or
cosa+sina=1
how to find a?
the only why which i came out with is to square, but i think if i do so i'll lose (not always but sometimes) several meanings of a..
i hope you understood what i was trying to say
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,...