Solving Trigonometric Equations for Angle Values

The problem provides two trigonometric equations involving two angles, x and y. We are given:

• $\cos (x−y) = \frac{\sqrt{3}}{2}$
• $\sin (x + y) = 1$

We are also given the condition that $x > y$. Our goal is to find the value of y.

Analyzing the Given Equations

Let's analyze each equation separately to find the possible values for $(x-y)$ and $(x+y)$.

Equation 1: $\cos (x−y) = \frac{\sqrt{3}}{2}$

We know that the cosine of certain standard angles is $\frac{\sqrt{3}}{2}$. The principal value for which cosine is $\frac{\sqrt{3}}{2}$ is $30°$.

So, we can write:

$x - y = 30°$

In general, $\cos \theta = \cos \alpha$ implies $\theta = 2n\pi \pm \alpha$. However, given the options and the typical context of such problems, we will consider the principal values, likely within $0^\circ$ to $180^\circ$ or $0^\circ$ to $360^\circ$ for the sums/differences. Since $x > y$, $x-y$ must be positive. Considering the most common range for angles in these problems ($0^\circ$ to $90^\circ$ for x and y), $x-y$ is likely $30°$.

Equation 2: $\sin (x + y) = 1$

We know that the sine of certain standard angles is 1. The principal value for which sine is 1 is $90°$.

So, we can write:

$x + y = 90°$

In general, $\sin \theta = \sin \alpha$ implies $\theta = n\pi + (-1)^n \alpha$. For $\alpha = 90°$, this gives $x+y = 90° + 360°n$. Again, considering the likely range for angles in this context, $x+y$ is likely $90°$.

Forming a System of Linear Equations

Now we have a system of two linear equations with two variables, x and y:

1. $x - y = 30°$
2. $x + y = 90°$

Solving the System for x and y

We can solve this system using the elimination method. Add the two equations together:

$(x - y) + (x + y) = 30° + 90°$

$2x = 120°$

Divide by 2 to find the value of x:

$x = \frac{120°}{2}$

$x = 60°$

Now substitute the value of x ($60°$) into the second equation ($x + y = 90°$) to find the value of y:

$60° + y = 90°$

Subtract $60°$ from both sides:

$y = 90° - 60°$

$y = 30°$

Verifying the Condition x > y

We found $x = 60°$ and $y = 30°$. Let's check if the condition $x > y$ is satisfied:

$60° > 30°$

This condition is true.

Final Answer

The value of y that satisfies the given conditions is $30°$.

Revision Table: Key Trigonometric Values

Additional Information: Trigonometric Equations and Principal Values

When solving trigonometric equations like $\cos \theta = k$ or $\sin \theta = k$, there are usually infinitely many solutions because trigonometric functions are periodic. However, in many high school math problems, particularly those with angle constraints or options suggesting acute angles, we focus on the principal values.

• For $\cos \theta = k$, the principal value is usually taken in the range $0° \le \theta \le 180°$ or $[0, \pi]$ radians. In this problem, $\cos(x-y) = \sqrt{3}/2$, the principal value for $x-y$ is $30°$.
• For $\sin \theta = k$, the principal value is usually taken in the range $-90° \le \theta \le 90°$ or $[-\pi/2, \pi/2]$ radians. In this problem, $\sin(x+y) = 1$, the principal value for $x+y$ is $90°$.

The system of equations derived from these principal values led to a consistent solution that also satisfied the given condition $x > y$. If a problem required considering other solutions (due to a wider domain for x and y), the general solutions would be used, leading to multiple possible pairs of (x, y) that satisfy the trigonometric equations, but only one pair would satisfy the $x > y$ condition and likely produce one of the options for y.