cos ^ 2 - sin ^ 2 =0?

Question

solve equation for solutions over the interval [0, 360)

Answer 1

sin^2 + cos^2 = 1
sin^2 = 1 - cos^2

sub in:

cos^2 - 1 + cos^2 = 0
2*cos^2 = 1
cos = 1/sqrt(2) or -1/sqrt(2)

cos(x) = +-1/sqrt(2)
cos(x) = +-[1/sqrt(2)]*[sqrt(2)*sqrt(2)]
cos(x) = +-sqrt(2)/2

x = 45, 135, 225, 315 deg

Answer 2

Visualization seems to be a lost art on this side
of the Atlantic; that's like doing math with one
hand tied behind your back.

Think of plots of cos x and sin x. (And, for goodness
sake, PLEASE show where x should go in your
original question.):

From x = 0, sin x starts up from 0 to 1 monotonically,
cos x starts down from 1 to 0 monotonically. There
can only be one place where they're equal, at

x = 45 deg. (where both = 2^[-1/2]).

Because the original trig fns. are squared, there's
a +/- sign ambiguity; so, 135 deg., 225 deg., and
315 deg satisfy it too.

Remember: there is more than one way to skin a
cat.

Live long and prosper.

Answer 3

Use the identity cos ^2 x - sin ^2 x cos 2x.
So cos 2x = 0,
2x = 90 or 270
x = 45 or 135. (All angles in degrees)

Answer 4

tan^2 = 1

Which gives pi/4, 3pi/4, 5pi/4 and 7pi/4 radians.