If A = cos²
θ + (sin²θ)² for all values of θ, then prove that
3
/4 ≤ A ≤ 1.
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Answered by
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.
=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.
Answered by
0
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.
=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.
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