Question:
If \( \cos x+\cos^{2}x=1 \) , then \( \sin^{6}x+3 \sin^{8}x+3 \sin^{10}x+\sin^{12}x= \)
If \( \cos x+\cos^{2}x=1 \) , then \( \sin^{6}x+3 \sin^{8}x+3 \sin^{10}x+\sin^{12}x= \)
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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!
Updated On: Oct 16, 2026
- \( 0 \)
- \( 1 \)
- \( 2 \)
- \( 3 \)
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The Correct Option is B
Solution and Explanation
Concept:
From the given trigonometric identity \( \cos x + \cos^2 x = 1 \), we can establish a direct link between cosine and sine functions:
\[
\cos x = 1 - \cos^2 x \implies \cos x = \sin^2 x
\]
This allows us to convert the high powers of sine in the target expression into lower powers of cosine.
Step 1: Recognizing a perfect cube structure in the target expression.
Let us look closely at the expression: \[ E = \sin^{6}x+3 \sin^{8}x+3 \sin^{10}x+\sin^{12}x \] Rearranging terms by ascending power order: \[ E = (\sin^2 x)^3 + 3(\sin^2 x)^2(\sin^4 x) + 3(\sin^2 x)(\sin^4 x)^2 + (\sin^4 x)^3 \] This perfectly matches the algebraic expansion for a perfect cube \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), where \( a = \sin^2 x \) and \( b = \sin^4 x \): \[ E = (\sin^2 x + \sin^4 x)^3 \]
Step 2: Substituting the cosine identity.
Since \( \sin^2 x = \cos x \), it follows that \( \sin^4 x = (\sin^2 x)^2 = \cos^2 x \). Let us substitute these into our simplified expression: \[ E = (\cos x + \cos^2 x)^3 \]
Step 3: Evaluating the final constant value.
We are given in the problem statement that \( \cos x + \cos^2 x = 1 \). Therefore: \[ E = (1)^3 = 1 \]
Step 1: Recognizing a perfect cube structure in the target expression.
Let us look closely at the expression: \[ E = \sin^{6}x+3 \sin^{8}x+3 \sin^{10}x+\sin^{12}x \] Rearranging terms by ascending power order: \[ E = (\sin^2 x)^3 + 3(\sin^2 x)^2(\sin^4 x) + 3(\sin^2 x)(\sin^4 x)^2 + (\sin^4 x)^3 \] This perfectly matches the algebraic expansion for a perfect cube \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), where \( a = \sin^2 x \) and \( b = \sin^4 x \): \[ E = (\sin^2 x + \sin^4 x)^3 \]
Step 2: Substituting the cosine identity.
Since \( \sin^2 x = \cos x \), it follows that \( \sin^4 x = (\sin^2 x)^2 = \cos^2 x \). Let us substitute these into our simplified expression: \[ E = (\cos x + \cos^2 x)^3 \]
Step 3: Evaluating the final constant value.
We are given in the problem statement that \( \cos x + \cos^2 x = 1 \). Therefore: \[ E = (1)^3 = 1 \]
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