Question

decide whether the following equation is a trigonometric identity. explain your reasoning.

cos^2(2x)-sin^2(2x)=0

Answer 1

start with double angle identity to get

cos (4x) = 0 .....(since cos 2x = cos^2 x - sin^2 x)

Answer 2

\[(1-\sin^2(2x))-\sin^2(2x)=1-2\sin^2(2x)=0\]

Answer 3

4x = pi/2 + k*pi (for any integer k)

x = pi/8 + k*pi/4 (for any integer k)

Answer 4

\[\sin^2(2x)=\frac{1}{2}\]

Answer 5

\[\sin(2x)=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]
sinu is sqrt{2}/2 when u =pi/4 and 3pi/4
and if we keep going around the circle
\[u=\frac{\pi}{4}+2n \pi, u=\frac{3\pi}{4}+2n \pi\]
so
\[2x=\frac{\pi}{4}+2n \pi, 2x=\frac{3\pi}{4}+2n \pi\]

Answer 6

\[x=\frac{\pi}{8}+n \pi, x=\frac{3\pi}{8}+npi\]

Answer 7

n is an integer

Answer 8

not a trig identity.
cos^2(x)-Sin^2(x)=cos(2x)

Answer 9

oops i didn't read lol

Answer 10

not 0. it is an equation