Question

Knowing that -1≤ cos(2x)≤ 1 , how do we know that 0≤ cos^2(2x)≤ 1

Answer 1

when you square a real number it is non-negative

also

\[\cos(2x)\le 1\Rightarrow \cos^2(2x)\le1^2=1\]

finally it is possible for \(\cos(2x)\) to be zero for example let \(x=\pi/4\)

thus

\[0\le\cos^2(2x)\le 1\]

Answer 2

if we square -1≤ cos(2x)≤ 1 , for all sides, the outcome is confusing for me

Answer 3

sorry to say but i feel sorry for you
you have so hard questions i dont understand
i wish i could help

Answer 4

if we square -1≤ cos(2x)≤ 1 , for all sides,what do we get?

Answer 5

graph each function to see what is going on.

ie graph \(\cos(2x)\) and \(\cos^2(2x)\)

Answer 6

how about algebraically

Answer 7

You can do it like this:

\[ \cos^2 (2x) = \frac {(1+\cos 4x)}{2} \]

Now,

\(-1 \le \cos 4x \le 1 \implies 0 \le 1+\cos 4x \le 2 \)\[\implies 0 \le \frac {(1+\cos 4x)}{2} \le 1 \implies 0\le \cos 2x \le 1\]

Answer 8

EDIT: Read the last line as

\[ \implies 0 \le \frac {(1+\cos 4x)}{2} \le 1 \implies 0\le \cos^2 (2x) \le 1 \]

Answer 9

when a cosine function is squared like cos^2x its graph is always +ve and thus its range is from 0<=cos^2x<=1.