If x sin3θ + y cos3θ = sin θ cos θ and x sin θ - y cos θ = 0, for every $\theta \in\left(0, \frac{\pi}{2}\right) $, then what is x2 + y2 equal to ?

Question

If x sin3&theta; + y cos3&theta; = sin &theta; cos &theta; and x sin &theta; - y cos &theta; = 0, for every $\theta \in\left(0, \frac{\pi}{2}\right) $, then what is x2 + y2 equal to ?

Prepp Team · Accepted Answer

Solving Trigonometric Equations for $x^2 + y^2$ We are given two trigonometric equations involving variables $x$ and $y$, and a trigonometric function of $\theta$. Our goal is to find the value of $x^2 + y^2$. The given equations are: $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ $x \sin \theta - y \cos \theta = 0$ The problem states that these equations hold for every $\theta \in \left(0, \frac{\pi}{2}\right)$. This interval is important because in this range, $\sin \theta > 0$ and $\cos \theta > 0$. Step 1: Analyze the Second Equation Let's start with the simpler second equation: $x \sin \theta - y \cos \theta = 0$ We can rearrange this equation to express one variable in terms of the other. Let's solve for $y$: $x \sin \theta = y \cos \theta$ Since $\theta \in \left(0, \frac{\pi}{2}\right)$, $\cos \theta \neq 0$. We can divide both sides by $\cos \theta$: $y = x \frac{\sin \theta}{\cos \theta}$ Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we get: $y = x \tan \theta$ Step 2: Substitute into the First Equation Now, substitute the expression for $y$ from Step 1 into the first equation: $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ Replace $y$ with $x \tan \theta$: $x \sin^3 \theta + (x \tan \theta) \cos^3 \theta = \sin \theta \cos \theta$ $x \sin^3 \theta + x \left(\frac{\sin \theta}{\cos \theta}\right) \cos^3 \theta = \sin \theta \cos \theta$ $x \sin^3 \theta + x \sin \theta \cos^2 \theta = \sin \theta \cos \theta$ Step 3: Simplify and Solve for $x$ Factor out the common terms on the left side, which are $x$ and $\sin \theta$: $x \sin \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta$ Recall the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$. Substitute this into the equation: $x \sin \theta (1) = \sin \theta \cos \theta$ $x \sin \theta = \sin \theta \cos \theta$ Since $\theta \in \left(0, \frac{\pi}{2}\right)$, $\sin \theta \neq 0$. We can divide both sides by $\sin \theta$: $x = \cos \theta$ Step 4: Solve for $y$ Now that we have the value of $x$, we can use the relationship $y = x \tan \theta$ from Step 1 to find $y$. Substitute $x = \cos \theta$: $y = (\cos \theta) \tan \theta$ $y = \cos \theta \left(\frac{\sin \theta}{\cos \theta}\right)$ Since $\theta \in \left(0, \frac{\pi}{2}\right)$, $\cos \theta \neq 0$. We can cancel out $\cos \theta$: $y = \sin \theta$ Step 5: Calculate $x^2 + y^2$ We have found that $x = \cos \theta$ and $y = \sin \theta$. Now we can calculate $x^2 + y^2$: $x^2 + y^2 = (\cos \theta)^2 + (\sin \theta)^2$ $x^2 + y^2 = \cos^2 \theta + \sin^2 \theta$ Using the fundamental trigonometric identity again, $\sin^2 \theta + \cos^2 \theta = 1$. Therefore: $x^2 + y^2 = 1$ The value of $x^2 + y^2$ is 1. Summary of Solutions for x and y Variable Value $x$ $\cos \theta$ $y$ $\sin \theta$ Checking the Solution with the Given Options The options provided are 0, 1, 2, and 3. Our calculated value for $x^2 + y^2$ is 1. This matches one of the options. Revision Table: Key Trigonometric Concepts Concept Description Identity/Property Tangent Identity Ratio of sine to cosine $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Pythagorean Identity Relationship between sine and cosine squared $\sin^2 \theta + \cos^2 \theta = 1$ Domain $\theta \in (0, \frac{\pi}{2})$ Specifies that $\theta$ is in the first quadrant $\sin \theta > 0$, $\cos \theta > 0$, $\tan \theta > 0$ Additional Information: System of Trigonometric Equations This problem involved solving a system of two equations where the variables $x$ and $y$ are linked through trigonometric functions of an angle $\theta$. The technique used was substitution, which is a common method for solving systems of equations. The fact that the equations hold for *every* $\theta \in \left(0, \frac{\pi}{2}\right)$ implies that the values of $x$ and $y$ must be constants that work for all valid $\theta$. By solving for $x$ and $y$ in terms of $\theta$, we found that $x$ and $y$ themselves depend on $\theta$. However, the final expression $x^2 + y^2$ simplified to a constant value, independent of $\theta$, which is the required answer. When solving trigonometric equations, it is crucial to pay attention to the domain of the angle $\theta$, as this can affect whether certain operations (like division by $\sin \theta$ or $\cos \theta$) are valid.

Prepp Team · Answer

0

Prepp Team · Answer

2

Prepp Team · Answer

3

Variable	Value
\(x\)	\(\cos \theta\)
\(y\)	\(\sin \theta\)

Solving Trigonometric Equations for \(x^2 + y^2\)

Step 1: Analyze the Second Equation

Step 2: Substitute into the First Equation

Step 3: Simplify and Solve for \(x\)

Step 4: Solve for \(y\)

Step 5: Calculate \(x^2 + y^2\)

Summary of Solutions for x and y

Checking the Solution with the Given Options

Revision Table: Key Trigonometric Concepts

Additional Information: System of Trigonometric Equations

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Concept	Description	Identity/Property
Tangent Identity	Ratio of sine to cosine	\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
Pythagorean Identity	Relationship between sine and cosine squared	\(\sin^2 \theta + \cos^2 \theta = 1\)
Domain \(\theta \in (0, \frac{\pi}{2})\)	Specifies that \(\theta\) is in the first quadrant	\(\sin \theta > 0\), \(\cos \theta > 0\), \(\tan \theta > 0\)