Question

The expression $3\Bigg[sin^4\Bigg(\frac{3\pi}{2}-\alpha\Bigg)+sin^4(3\pi+\alpha)\Bigg]-2\Bigg[sin^6\Bigg(\frac{\pi}{2}+\alpha\Bigg)+sin^6(5\pi-\alpha)\Bigg]$ is equal to

• A. 0
• B. 1
• C. 3
• D. $sin 4\alpha + cos 6 \alpha$

Answer 1

Answer: (B)

1

Answer 2

The correct answer is B:1
$3\Bigg[sin^4\Bigg(\frac{3\pi}{2}-\alpha\Bigg)+sin^4 (3\pi+\alpha)\Bigg]-2\Bigg[sin^6\Big(\frac{\pi}{2}+\alpha\Big)+sin^6 (5\pi-\alpha)]$
$= 3 (cos^4 \alpha + sin^4 \alpha )-2 (cos^6 \alpha + sin^6 \alpha )$
$=3[(cos^2x)^2+(sin^2x)^2]-2[(sin^2x)^3+(cos^2x)^3]$
$=2[(sin^2x+cos^2x)(sin^4x+cos^4x-sin^2x.cos^2x)]$
We know that ,$a^4+b^4=(a^2+b^2)-2a^2b^2$
and$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)(a^4+b^4-a^2b^2)$
From the above form we can write;
$= 3 (1 - 2 sin^2 \alpha cos^2 \alpha ) -2 (1 - 3 sin^2 \alpha cos^2 \alpha)$
$= 3 - 6 sin^2 \alpha cos^2 \alpha - 2 + 6 sin^2 \alpha cos^2 \alpha = 1$