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Chapter 5: Problem 16

Verify the identity. $$\sin ^{2} \alpha-\sin ^{4} \alpha=\cos ^{2} \alpha-\cos ^{4} \alpha$$

Short Answer

Expert verified
The given equation is indeed an identity. This was achieved by converting both sides into a difference of squares, which revealed the identical structure on both sides of the equation.

Step by step solution

01

Split Each Side Into Two Squares

We start by writing each side as a difference of two squares. This gives: \[ \sin^2 \alpha - \sin^4 \alpha = (\sin^2 \alpha - \sin^2 \alpha)(\sin^2 \alpha + \sin^2 \alpha) \] and similarly for \(\cos\), \[ \cos^2 \alpha - \cos^4 \alpha = (\cos^2 \alpha - 1)(\cos^2 \alpha + 1) \]
02

Simplify Each Side

Simplify the expressions: For \(\sin\), because \(\sin^2 \alpha + \sin^2 \alpha = 1 - \cos^2 \alpha\), we have \[ (\sin^2 \alpha - \sin^2 \alpha)(1 - \cos^2 \alpha)\]. Similarly for \(\cos\), \(\cos^2 \alpha + 1 = 1 + \cos^2 \alpha = 1 - sin^2 \alpha\), giving us \[ (\cos^2 \alpha - 1)(1 - \sin^2 \alpha)\].
03

Make the Conclusion

The two sides match since \(\sin^2 \alpha - \sin^4 \alpha = (\sin ^2 \alpha)(1 - \cos^2 \alpha)\) and \(\cos^2 \alpha - \cos^4 \alpha = (\cos ^2 \alpha)(1 - \sin^2 \alpha)\), hence the given equation is an identity.

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