Question

If \( \cos x+\cos^{2}x=1 \) , then \( \sin^{6}x+3 \sin^{8}x+3 \sin^{10}x+\sin^{12}x= \)

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

Whenever you see a polynomial with coefficients following the pattern \( 1, 3, 3, 1 \), it is a dead giveaway for a perfect cube expansion \( (a+b)^3 \). Identifying this pattern saves you from doing long conversions term by term!

Answer 1

The Correct Option is B

Step 1: Use the given relation.
From $\cos x + \cos^2 x = 1$, move terms to get $\cos x = 1 - \cos^2 x$. The right side is $\sin^2 x$, so $\cos x = \sin^2 x$.
Step 2: Look at the target.
We want $E = \sin^6 x + 3\sin^8 x + 3\sin^{10} x + \sin^{12} x$.
Step 3: Spot a cube pattern.
Write each term using $a = \sin^2 x$ and $b = \sin^4 x$. Then $E = a^3 + 3a^2b + 3ab^2 + b^3 = (a+b)^3$.
Step 4: Replace a and b.
So $E = (\sin^2 x + \sin^4 x)^3$. Since $\sin^2 x = \cos x$, we get $\sin^4 x = \cos^2 x$.
Step 5: Substitute back.
Thus $E = (\cos x + \cos^2 x)^3$.
Step 6: Use the given value.
But $\cos x + \cos^2 x = 1$, so $E = 1^3$. \[ \boxed{1} \]