Note that the RHS simplifies to cos²x.

In short, we try to prove that
|cos x| ≥ cos²x
for all real values of x.

Now, we know that cos x has a lowest possible value of -1 and a greatest possible value of 1, such that -1 ≤ cos x ≤ 1.

A number being squared means that the number is being multiplied by itself. We know that multiplying a positive number with a positive fraction below 1 will reduce the value of the positive number. So, if the number being squared is less than 1, then the number is essentially multiplied by another of itself which is less than 1, so the value of the number decreases.

It is, in fact, true that for -1 ≤ u ≤ 1. u² ≤ |u|.

For the positive (or at least non-negative) case 0 ≤ cos x ≤ 1, this is easy, as per explained above.

The negative case -1 ≤ cos x < 0 is basically identical to the positive case, since cos²x and |cos x| is symmetric about the vertical axis (these are properties of squared and modulused expressions).

As such, |cos x| ≥ 1 - sin²x for all x E R.

Thank you.

-1 ≤ cos x ≤ 1 for all x ∈ R

|cos x| = -cos x when -1 ≤ cos x < 0,
such that 0 < -cos x ≤ 1

|cos x| = cos x when 0 ≤ cos x ≤ 1

Therefore, 0 ≤|cos x| ≤ 1 for all x ∈ R


Next,
1 - sin² x = cos² x

When 0 ≤|cos x|≤ 1 for all x ∈ R,
0 ≤|cos x|² ≤ 1 for all x ∈ R

Since |cos x|² = cos² x for all x ∈ R,
0 ≤ cos² x ≤ 1 for all x ∈ R



①
When cos x = -1, 0 or 1,

|cos x| = 1 ,0 or 1 and cos² x = 1, 0 or 1

So |cos x| = cos² x = 1 - sin² x when cos x = -1,0,1



②
For 0 < cos x < 1,

cos x > cos² x

Since |cos x| = cos x for this range,
then |cos x|> cos² x for this range.

Which means |cos x|> 1 - sin² x for this range


③
For -1 < cos x < 0,

0 < -cos x < 1

-cos x > (-cos x)²
-cos x > cos² x

Since |cos x| = -cos x for this range,
then |cos x|> cos² x for this range.

Which means |cos x|> 1 - sin² x for this range.


Therefore, ∣ cos x|> 1 - sin² x for all x ∈ R

Edit : last step the inequality should be ≥