Trigonometrical Ratios and Identities 1 Question 12
12. $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)$ equals
$(1995,2 M)$
(a) 11
(b) 12
(c) 13
(d) 14
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Answer:
Correct Answer: 12. (c)
Solution:
- Given expression $=$
$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)$
$=3(1-\sin 2 x)^{2}+6(1+\sin 2 x)+4{\left(\sin ^{2} x+\cos ^{2} x\right)^{3}\right.$ $\left.-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right) }$
$=3\left(1-2 \sin 2 x+\sin ^{2} 2 x\right)+6+6 \sin 2 x$
$$ +4\left(1-3 \sin ^{2} x \cos ^{2} x\right) $$
$=3\left(1-2 \sin 2 x+\sin ^{2} 2 x+2+2 \sin 2 x\right)+4 \quad 1-\frac{3}{4} \cdot \sin ^{2} 2 x$
$=13+3 \sin ^{2} 2 x-3 \sin ^{2} 2 x=13$
