if A=cos^2 theta + sin^4 theta, then prove that for all values of theta, 3/4 less than equal to A less than 1?

Answer 1

Part I
cos^2X + sin^4X = 1 - sin^2X + sin^4x
= 1 + (sin^2X -1/2)^2 -1/4 >= 3/4 which is obvious, square of any number is always > 0
Part II
cos^2X + sin^4x <= cos^2X + sin^2X = 1, since (sin^2X < =1)
Combining
3/4 <= cos^2X + sin ^4X <= 1, isn't it.

Answer 2

A = cos^2(theta) + sin^4(theta)

Prove: 3/4 <= A < 1

Impossible to prove. If theta is 0, then cos^2(theta) is 1 and sin^4(theta) is 0. Added up is 1, which is outside the range.

Answer 3

3/4 <= A < 1

A = cos^2theta + sin^4 theta

A = cos^2theta + sin^4theta

A = 1 - sin^2theta + sin^4theta

A = 1 - sin^2theta*(1-sin^2theta)

A = 1 - sin^2theta*cos^2theta

(sin^2theta + cos^2theta)^2 = sin^4theta + cos^4theta +2sin^2theta*cos^2theta

then : 1 - 2sin^theta*cos^2theta = sin^4theta + cos^4theta

A = 1 - sin^2theta*cos^2theta

2A = 1 + 1 - 2sin^2theta*cos^2theta

2A = 1 + (sin^4theta + cos^4theta)

But, here there is a useful property :

sin^4theta + cos^4theta = 3/4 + 1/4(sin(4theta))

then :

2A = 1 + 3/4 + 1/4(sin(4theta))

A = ( 1 + 3/4 + 1/4(sin(4theta)) / 2

-1<=sin(4theta) <=1

-1/4<= 1/4*sin(4theta) <= 1/4

3/2 <= 1 + 3/4 + 1/4*sin(4theta) <= 2

3/4 <= A <= 1

That's it

Hope that might help you

Answer 4

Sounds like highschool math. You are not gettin anywhere this way. Do you own home work.