1. The problem is to verify the identity: $\sin^4 b + \cos^4 b = 1 - \frac{1}{2} \sin^2 2b$. 2. Recall the Pythagorean identity: $\sin^2 b + \cos^2 b = 1$. 3. Use the formula for the square of a sum: $a^2 + b^2 = (a+b)^2 - 2ab$. 4. Rewrite the left side: $$\sin^4 b + \cos^4 b = (\sin^2 b)^2 + (\cos^2 b)^2 = (\sin^2 b + \cos^2 b)^2 - 2 \sin^2 b \cos^2 b$$ 5. Substitute $\sin^2 b + \cos^2 b = 1$: $$= 1^2 - 2 \sin^2 b \cos^2 b = 1 - 2 \sin^2 b \cos^2 b$$ 6. Use the double angle identity for sine: $$\sin 2b = 2 \sin b \cos b \implies \sin^2 2b = 4 \sin^2 b \cos^2 b$$ 7. Substitute $\sin^2 b \cos^2 b = \frac{1}{4} \sin^2 2b$ into the expression: $$1 - 2 \times \frac{1}{4} \sin^2 2b = 1 - \frac{1}{2} \sin^2 2b$$ 8. This matches the right side of the identity, so the identity is verified. Final answer: $\sin^4 b + \cos^4 b = 1 - \frac{1}{2} \sin^2 2b$ is true.