If asin³θ  + bcos³θ = sinθ cosθ and asinθ = bcosθ prove that a²+b²= 1

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 \( a^2 + b^2 = 1 \).

---

### 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 \( a \):

\[
b = a \frac{\sin \theta}{\cos \theta}
\]

---

### 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:

\[
a^2 \sin^2 \theta - b^2 \cos^2 \theta = 0
\]

---

### 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.

---

### 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.

If asin³θ  + bcos³θ = sinθ cosθ and asinθ = bcosθ prove that a² + b² = 1

Discover how to prove that a² + b² = 1 with trigonometric identities and algebraic manipulation, given the equations asin³θ  + bcos³θ = sinθ cosθ and asinθ = bcosθ. Ideal for high school students (Grades 10-12) studying trigonometry.

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