Solving the Trigonometric Equation and Evaluating cos³x + cos³y

The problem provides a trigonometric equation \(\sin x + \sin y = \sqrt{3} (\cos y - \cos x)\) and a relation \(x + y = \frac{\pi}{2}\). We are asked to find the value of \(\cos^3 x + \cos^3 y\). The conditions on the range of x and y, stated as "x < x, y \(\rm \frac{\pi}{2}\)", are unclear. We will proceed by using the given equation and \(x+y = \frac{\pi}{2}\) to find suitable values for \(x\) and \(y\) that lead to one of the provided options.

Applying Trigonometric Identities

We use the sum-to-product and difference-to-product identities:

• \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)
• \(\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\)

Applying these to the given equation \(\sin x + \sin y = \sqrt{3} (\cos y - \cos x)\):

\(2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = \sqrt{3} \left(-2 \sin\left(\frac{y+x}{2}\right) \sin\left(\frac{y-x}{2}\right)\right)\)

Given \(x + y = \frac{\pi}{2}\), we have \(\frac{x+y}{2} = \frac{\pi}{4}\). Also, \(\sin\left(\frac{y-x}{2}\right) = -\sin\left(\frac{x-y}{2}\right)\). Substitute these into the equation:

\(2 \sin\left(\frac{\pi}{4}\right) \cos\left(\frac{x-y}{2}\right) = \sqrt{3} \left(-2 \sin\left(\frac{\pi}{4}\right) \left(-\sin\left(\frac{x-y}{2}\right)\right)\right)\)

\(2 \sin\left(\frac{\pi}{4}\right) \cos\left(\frac{x-y}{2}\right) = \sqrt{3} \left(2 \sin\left(\frac{\pi}{4}\right) \sin\left(\frac{x-y}{2}\right)\right)\)

Since \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \neq 0\), we can divide both sides by \(2 \sin\left(\frac{\pi}{4}\right)\):

\(\cos\left(\frac{x-y}{2}\right) = \sqrt{3} \sin\left(\frac{x-y}{2}\right)\)

Assuming \(\cos\left(\frac{x-y}{2}\right) \neq 0\), we divide by it:

\(\frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} = \frac{1}{\sqrt{3}}\)

\(\tan\left(\frac{x-y}{2}\right) = \frac{1}{\sqrt{3}}\)

Determining Values for x and y

The general solution for \(\tan \theta = \frac{1}{\sqrt{3}}\) is \(\theta = \frac{\pi}{6} + n\pi\), where \(n\) is an integer.

So, \(\frac{x-y}{2} = \frac{\pi}{6} + n\pi\). This implies \(x-y = \frac{\pi}{3} + 2n\pi\).

We also have the condition \(x+y = \frac{\pi}{2}\). We can solve this system of linear equations for \(x\) and \(y\):

Adding the two equations: \(2x = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi = \frac{5\pi}{6} + 2n\pi \implies x = \frac{5\pi}{12} + n\pi\).

Subtracting the second equation from the first: \(2y = \frac{\pi}{2} - \left(\frac{\pi}{3} + 2n\pi\right) = \frac{\pi}{6} - 2n\pi \implies y = \frac{\pi}{12} - n\pi\).

Possible pairs \((x,y)\) are \((\frac{5\pi}{12} + n\pi, \frac{\pi}{12} - n\pi)\) for integer \(n\). Considering common angle ranges in such problems, let's evaluate the expression for \(n=0\), which gives \(x = \frac{5\pi}{12}\) and \(y = \frac{\pi}{12}\). Both angles are positive and less than \(\frac{\pi}{2}\). Let's calculate \(\cos^3 x + \cos^3 y\) for this pair.

Calculating cos&sup3;x + cos&sup3;y for x = 5π/12 and y = π/12

We find the cosine values for \(x\) and \(y\):

\(\cos x = \cos\left(\frac{5\pi}{12}\right) = \cos(75^\circ)\). Using the angle addition formula \(\cos(A+B) = \cos A \cos B - \sin A \sin B\) with \(75^\circ = 45^\circ + 30^\circ\):

\(\cos(75^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}\).

\(\cos y = \cos\left(\frac{\pi}{12}\right) = \cos(15^\circ)\). Using the angle subtraction formula \(\cos(A-B) = \cos A \cos B + \sin A \sin B\) with \(15^\circ = 45^\circ - 30^\circ\):

\(\cos(15^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}\).

Now we calculate \(\cos^3 x + \cos^3 y\):

\(\cos^3 x + \cos^3 y = \left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^3 + \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^3\)

Let \(a = \frac{\sqrt{6}}{4}\) and \(b = \frac{\sqrt{2}}{4}\). The expression is in the form \((a-b)^3 + (a+b)^3\). Using the algebraic identity \((a-b)^3 + (a+b)^3 = (a^3 - 3a^2b + 3ab^2 - b^3) + (a^3 + 3a^2b + 3ab^2 + b^3) = 2a^3 + 6ab^2 = 2a(a^2 + 3b^2)\).

First, find \(a^2\) and \(b^2\):

\(a^2 = \left(\frac{\sqrt{6}}{4}\right)^2 = \frac{6}{16} = \frac{3}{8}\)

\(b^2 = \left(\frac{\sqrt{2}}{4}\right)^2 = \frac{2}{16} = \frac{1}{8}\)

Substitute these values into the simplified expression \(2a(a^2 + 3b^2)\):

\(2\left(\frac{\sqrt{6}}{4}\right)\left(\frac{3}{8} + 3\left(\frac{1}{8}\right)\right) = \frac{\sqrt{6}}{2}\left(\frac{3}{8} + \frac{3}{8}\right) = \frac{\sqrt{6}}{2}\left(\frac{6}{8}\right)\)

\(\frac{\sqrt{6}}{2}\left(\frac{6}{8}\right) = \frac{\sqrt{6}}{2}\left(\frac{3}{4}\right) = \frac{3\sqrt{6}}{8}\)

Conclusion

For the values \(x = \frac{5\pi}{12}\) and \(y = \frac{\pi}{12}\), which satisfy \(x+y=\frac{\pi}{2}\) and result from the derived relationship \(\tan\left(\frac{x-y}{2}\right) = \frac{1}{\sqrt{3}}\) with \(n=0\), the value of \(\cos^3 x + \cos^3 y\) is \(\frac{3\sqrt{6}}{8}\). This matches one of the given options.