The value of 3 (sin x – cos x)

Question:

The value of 3 (sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6x + cos6x) is ___________

Solution:

3 (sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6x + cos6x)

$=3\left[(\sin x-\cos x)^{2}\right]^{2}+6\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)+4\left[\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right]$

 

$=3\left[\sin ^{2} x+\cos ^{2} x-2 \cos x \sin x\right]^{2}+6(1+2 \sin x \cos x)$

$+4\left[\left(\sin ^{2} x+\cos _{1}^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)\right]=1$

$=3[1-2 \cos x \sin x]^{2}+6+12 \sin x \cos x+4\left[1-3 \sin ^{2} x \cos ^{2} x\right]$

 

$=3\left[1+4 \cos ^{2} x \sin ^{2} x-4 \cos x \sin x\right]$

$+6+12 \sin x \cos x+4-12 \sin ^{2} x \cos ^{2} x$

$=3+12 \cos ^{2} x \sin ^{2} x-12 \cos x \sin x$

 

$+6+12 \cos x \sin x+4-12 \sin ^{2} x \cos ^{2} x$

= 13

Hence, $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)$ has value 13

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