For any (4,2), the expression 3(-)^4+6(+)^2+4^6 equals:

Question

For any $	heta\in\left(\dfrac\pi4,\dfrac\pi2ight)$, the expression $3(\sin	heta-\cos	heta)^4+6(\sin	heta+\cos	heta)^2+4\sin^6	heta$ equals:

Accepted Answer

Given,

3(sin$$\theta $$ - cos$$\theta $$)4 + 6(sin$$\theta $$ + cos$$\theta $$)2 + 4sin6$$\theta $$

=  3[(sin$$\theta $$ - cos$$\theta $$)2]2 + 6 (sin2$$\theta $$ + cos2$$\theta $$ + 2sin$$\theta $$cos$$\theta $$) + 4sin6$$\theta $$

=  3[sin2$$\theta $$ + cos2$$\theta $$ -2sin$$\theta $$cos$$\theta $$]2 + 6(1 + sin2$$\theta $$) + 4sin6$$\theta $$

=  3(1 - sin2$$\theta $$)2 + 6(1 + sin2$$\theta $$) + 4sin6$$\theta $$

=  3 (1 - 2 sin2$$\theta $$ + sin22$$\theta $$) + 6 + 6sin2$$\theta $$ + 4sin6$$\theta $$

=  3 - 6sin2$$\theta $$ + 3sin22$$\theta $$ + 6 + 6sin2$$\theta $$ + 4sin6$$\theta $$

=  9 + 3sin22$$\theta $$ + 4 sin6$$\theta $$

=  9 + 3(2sin$$\theta $$cos$$\theta $$)2 + 4(1 - cos2$$\theta $$)3

= 9 + 12sin2$$\theta $$ cos2$$\theta $$ + 4 (1 - cos6$$\theta $$ - 3cos2$$\theta $$ + 3cos4$$\theta $$)

=  13 + 12 (1 - cos2$$\theta $$ - 4cos6$$\theta $$ - 12cos$$\theta $$ + 12 cos4$$\theta $$

= 13 + 12 cos2$$\theta $$ - 12 cos4$$\theta $$ - 4cos6$$\theta $$ - 12 cos2$$\theta $$ + 12 cos4$$\theta $$

= 13 - 4 cos6$$\theta $$

Solutions · 1