Answer to Question #128122 in Trigonometry for Tony

Question

Prove that the inequality | cos(x)| ≥ 1 − sin2(x) holds true for all x ∈ R.

Answer 1

Removing the mode;

cos x ≤ - (1 - sin²x) and cos x ≥ (1 - sin²x)

Considering;

cos x ≤ - (1 - sin²x)

cos x ≤ - 1 + sin²x

cos x - sin²x + 1 ≤ 0

cos x - (1 - cos²x) + 1 ≤ 0

cos x + cos²x ≤ 0

-1 ≤ cos x ≤ 0

True for all x ∈ R  and π/2+2πn ≤ x ≤ 3π/2+2πn .......(i)

Considering;

cos x ≥  (1 - sin²x)

cos x ≥  1 - sin²x

cos x + sin²x - 1 ≥ 0

cos x + (1 - cos²x) - 1 ≥ 0

cos x - cos²x + 1 ≥ 1

0 ≤ cos x ≤ 1

Therefore for all x ∈ R  and -π/2+2πn ≤ x ≤ π/2+2πn .......(ii)

combining equations (i) and (ii) we get,

For all x ∈ R  and -π/2+2πn ≤ x ≤ 3π/2+2πn.