Question

If \( \sin x + \sin^2 x = 1 \) then \( \cos^8 x + 2\cos^6 x + \cos^4 x \) is equal to:

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

Hint:

Whenever powers of \( \cos x \) appear, try expressing everything in terms of \( \sin x \) using identitiesThen use given equations to simplify quickly.

Answer 1

The Correct Option is B

Step 1: Let \( \sin x = t \).
Given equation becomes:
\[ t + t^2 = 1 \]

Step 2: Rearrange the equation.
\[ t^2 + t - 1 = 0 \]

Step 3: Express \( \cos^2 x \) in terms of \( t \).
\[ \cos^2 x = 1 - \sin^2 x = 1 - t^2 \]
From
Step 1:
\[ t^2 = 1 - t \]
So:
\[ \cos^2 x = 1 - (1 - t) = t \]

Step 4: Simplify the given expression.
\[ \cos^8 x + 2\cos^6 x + \cos^4 x = (\cos^4 x)^2 + 2(\cos^4 x)(\cos^2 x) + (\cos^2 x)^2 \]
\[ = (\cos^4 x + \cos^2 x)^2 \]

Step 5: Substitute \( \cos^2 x = t \).
\[ \cos^4 x = t^2 \]
So expression becomes:
\[ (t^2 + t)^2 \]

Step 6: Use given relation.
From
Step 1:
\[ t^2 + t = 1 \]
Thus:
\[ (t^2 + t)^2 = 1^2 = 1 \]

Step 7: Final conclusion.
\[ \boxed{1} \]