The value of `3(cos theta-sin theta)^(4)+6(sin theta+cos theta)^(2)+4 sin^(6) theta` is where `theta in ((pi)/(4),(pi)/(2))`
(a) `13-4cos^(4) theta`
(b) `13-4cos^(6) theta`
(c) `13-4cos^(6) theta+ 2 sin^(4) theta cos^(2) theta`
(d) `13-4cos^(4) theta+ 2 sin^(4) theta cos^(2) theta`

Answer

D

Explanation

Given expression
`=3(sin theta - cos theta)^(4) + 6 (sin theta + cos theta)^(2) + 4sin^(6) theta`
`=3((sin theta - cos theta)^(2))^(2)+6(sin theta + cos theta)^(2)+4(sin^(2)theta)^(2)`
`=3 (1-sin 2 theta)^(2) + 6(1 + sin 2 theta) + 4 (1- cos^(2) theta)^(3) " "[because 1 + sin 2 theta = (cos theta + sin theta)^(2) and 1 - sin 2 theta = (cos theta - sin theta)^(2)]`
`=3(1^(2)+sin^(2) 2 theta - 2 sin 2 theta) + 6(1 + sin 2 theta)+4(1-cos^(6) theta - 3 cos^(2) theta + 3 cos^(4) theta) " "[because (a-b)^(2) = a^(2)+b^(2)-2ab and (a-b)^(3) = a^(3)-b^(3)-3a^(2)b+3ab^(2)]`
`=3 + 3 sin^(2) 2 theta - 6 sin 2 theta + 6 + 6 sin 2 theta + 4-4cos^(6) theta - 12 cos^(2) theta + 12 cos^(4) theta`
`=13+3sin^(2)2theta - 4 cos^(6) theta - 12 cos^(2) theta+ 12cos^(4) theta`
`=13+3(2sin theta cos theta)^(2) - 4 cos^(6) theta - 12 cos^(12) theta(1-cos^(2) theta)`
`=13 + 12 sin^(2)theta cos^(2)theta - 4cos^(6) theta - 12 cos^(2) theta sin^(2) theta`
`=13 - 4 cos^(6) theta`