JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 3)

For any $$\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)$$, the expression

$$3{(\cos \theta - \sin \theta )^4}$$$$ + 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta $$

equals :
13 – 4 cos2$$\theta $$ + 6sin2$$\theta $$cos2$$\theta $$
13 – 4 cos6$$\theta $$
13 – 4 cos2$$\theta $$ + 6cos2$$\theta $$
13 – 4 cos4$$\theta $$ + 2sin2$$\theta $$cos2$$\theta $$

Explanation

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$$\theta $$

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