\( \Large 3 \left(\sin x - \cos x\right)^{4} + 6 \left(\sin x + \cos x\right)^{2} + 4 \left(\sin^{6}x + \cos^{6}x\right) \) is equal to:


A) 11

B) 12

C) 13

D) 14

Correct Answer:
C) 13

Description for Correct answer:

\( \Large 3 \left(\sin x - \cos x\right)^{4}+6 \left(\sin x + \cos x\right)^{2} + 4 \left(\sin^{6}x+\cos^{6}x\right) \)

= \( \Large 3 \left(1- \sin 2x\right)^{2} + 6 \left(1+ \sin 2x\right) \) \( \Large + 4 \{ \left(\sin^{2}x+\cos^{2}x\right)^{3} \) \( \Large -3 \sin^{2}x \cos^{2}x \left(\sin^{2}x+\cos^{2}x\right) \}\)

 \(= \Large 3 \left(1-\sin 2x + \sin^{2}2x\right)+6+6 \sin 2x + 4\{ 1-3 \sin^{2}x \cos^{2}x \} \)

= \( \Large 3 \left(1-2 \sin 2x + \sin^{2}2x+2+2 \sin 2x\right)+4\{ 1 - \frac{3}{4}\sin^{2} 2x \} \)

\( \Large 13 + 3 \sin^{2} 2x - 3 \sin^{2} 2x = 13 \)


Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio








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