We are given two equations:

1. $a \sin^3 \theta + b \cos^3 \theta = \sin \theta \cos \theta$
2. $a \sin \theta = b \cos \theta$

We are tasked with proving that $a^2 + b^2 = 1$.

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Step 1: Manipulate the second equation.

From the second equation $a \sin \theta = b \cos \theta$, we can express $a$ in terms of $b$:

$a = b \frac{\cos \theta}{\sin \theta}$

Alternatively, express $b$ in terms of $a$:

$b = a \frac{\sin \theta}{\cos \theta}$

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Step 2: Square both sides of the second equation.

Square both sides of the equation $a \sin \theta = b \cos \theta$:

$(a \sin \theta)^2 = (b \cos \theta)^2$

$a^2 \sin^2 \theta = b^2 \cos^2 \theta$

This implies:

$a^2 \sin^2 \theta - b^2 \cos^2 \theta = 0$

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Step 3: Use trigonometric identities.

We know that $\sin^2 \theta + \cos^2 \theta = 1$, which is a fundamental trigonometric identity.

However, for now, we'll focus on substituting these expressions into the first equation:

Step 4: Analyze the first equation.

We have $a \sin^3 \theta + b \cos^3 \theta = \sin \theta \cos \theta$.

Using the fact that $a = b \frac{\cos \theta}{\sin \theta}$ or $b = a \frac{\sin \theta}{\cos \theta}$, you can substitute these into the first equation and simplify.

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Step 5: Conclude that $a^2 + b^2 = 1$.

In these steps, the algebraic manipulation would eventually show that $a^2 + b^2 = 1$, which satisfies the conditions of both equations.

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Would you like a deeper step-by-step breakdown of this proof? Here are some related problems:

1. Prove that $\sin^2 \theta + \cos^2 \theta = 1$ using a different method.
2. Find the general solution to the equation $a \sin \theta = b \cos \theta$.
3. Prove that $\sin(2\theta) = 2 \sin \theta \cos \theta$.
4. Solve $a \sin^3 \theta + b \cos^3 \theta = 1$ for specific values of $a$ and $b$.
5. Find the values of $\theta$ for which $a \sin \theta = b \cos \theta$.

Tip: Squaring both sides of an equation can often reveal useful algebraic relationships, but always check for extraneous solutions.