Verify the identity.
sin2A cos2A = Sin2A – 4sin3A cosA
The correct answer and explanation is :
To verify the identity:
[
\sin2A \cos2A = \sin2A – 4\sin^3A \cosA
]
We’ll verify if the left-hand side (LHS) equals the right-hand side (RHS) using trigonometric identities.
Step 1: Expand both sides using identities
Left-hand side (LHS):
[
\sin2A \cos2A
]
Use the double-angle identities:
- (\sin2A = 2\sin A \cos A)
- (\cos2A = \cos^2 A – \sin^2 A)
So,
[
\sin2A \cos2A = (2\sin A \cos A)(\cos^2 A – \sin^2 A)
]
Now multiply:
[
= 2\sin A \cos A (\cos^2 A – \sin^2 A)
]
Right-hand side (RHS):
[
\sin2A – 4\sin^3 A \cos A
]
Use:
(\sin2A = 2\sin A \cos A)
So:
[
\sin2A – 4\sin^3 A \cos A = 2\sin A \cos A – 4\sin^3 A \cos A
]
Factor:
[
= 2\sin A \cos A (1 – 2\sin^2 A)
]
Step 2: Compare LHS and RHS
We have:
- LHS: (2\sin A \cos A (\cos^2 A – \sin^2 A))
- RHS: (2\sin A \cos A (1 – 2\sin^2 A))
We now show that:
[
\cos^2 A – \sin^2 A = 1 – 2\sin^2 A
]
This is a known identity:
[
\cos 2A = \cos^2 A – \sin^2 A = 1 – 2\sin^2 A
]
✅ Therefore:
[
\cos^2 A – \sin^2 A = 1 – 2\sin^2 A
]
So both sides simplify to the same expression:
[
2\sin A \cos A (1 – 2\sin^2 A)
]
✅ Final Answer:
[
\sin2A \cos2A = \sin2A – 4\sin^3A \cosA \quad \text{is a TRUE identity.}
]
🔍 Summary Explanation (300+ words):
This identity can be verified using standard trigonometric identities, particularly the double-angle formulas for sine and cosine. The double-angle identity for sine is (\sin2A = 2\sin A \cos A), while for cosine there are multiple forms, including (\cos2A = \cos^2 A – \sin^2 A) and (\cos2A = 1 – 2\sin^2 A). These identities are key to rewriting complex expressions involving angles like (2A) or higher powers of sine and cosine.
In this problem, we start by simplifying both sides independently. For the left-hand side (LHS), we expand (\sin2A \cos2A) by substituting the known identities. The resulting expression becomes (2\sin A \cos A (\cos^2 A – \sin^2 A)). Meanwhile, the right-hand side (RHS), (\sin2A – 4\sin^3 A \cos A), can also be rewritten using the identity for (\sin2A). After factoring, it becomes (2\sin A \cos A (1 – 2\sin^2 A)).
To complete the verification, we show that the expressions inside the parentheses on both sides are equivalent. That is, we show that (\cos^2 A – \sin^2 A = 1 – 2\sin^2 A), which is a known identity for (\cos2A). Because both LHS and RHS simplify to the same exact expression, the identity is proven to be true.