Question

If A = cos²
θ + (sin²θ)² for all values of θ, then prove that
3
/4 ≤ A ≤ 1.

Answer 1

A= cos^2 x + sin^4 x
=cos^2 x(cos^2 x + sin^2 x) + sin^4 x
=(cos^4 x + sin^4 x + 2 cos^2 x sin^2 x) - cos^2 xsin^2 x
= (cos^2 x + sin^2 x)^2 - cos^2 x sin^2 x
= 1 - cos^2 x sin^2 x
= 1 - 1/4 (2 cos x sin x)^2
= 1 - 1/4 sin^2 (2x)

Now -1 <= sin theta <= 1, so 0<= sin^2 theta <=1
Hence A can take values between 3/4 to 1,  both inclusive.

Answer 2

A= cos^2 x + sin^4 x
=cos^2 x(cos^2 x + sin^2 x) + sin^4 x
=(cos^4 x + sin^4 x + 2 cos^2 x sin^2 x) - cos^2 xsin^2 x
= (cos^2 x + sin^2 x)^2 - cos^2 x sin^2 x
= 1 - cos^2 x sin^2 x
= 1 - 1/4 (2 cos x sin x)^2
= 1 - 1/4 sin^2 (2x)

Now -1 <= sin theta <= 1, so 0<= sin^2 theta <=1
Hence A can take values between 3/4 to 1, both inclusive.