Question

Evaluate : \(\frac{\sin^3 60^{\circ} - \tan 30^{\circ}}{\cos^2 45^{\circ}}\)

Hint:

Remember that \(\sin^3 \theta\) means \((\sin \theta)^3\). When simplifying fractions with surds, it is often helpful to rationalize the final answer.

Answer 1

Step 1: Understanding the Concept:
This problem requires substituting the standard trigonometric values for \(60^{\circ}\), \(30^{\circ}\), and \(45^{\circ}\) and simplifying the resulting fraction.
Step 2: Key Formula or Approach:
Standard values:
\(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\)
\(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\)
\(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\)
Step 3: Detailed Explanation:
Numerator:
\[ \sin^3 60^{\circ} - \tan 30^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^3 - \frac{1}{\sqrt{3}} \]
\[ = \frac{3\sqrt{3}}{8} - \frac{1}{\sqrt{3}} \]
Take the L.C.M. of the denominators (\(8\) and \(\sqrt{3}\)):
\[ = \frac{3\sqrt{3}(\sqrt{3}) - 1(8)}{8\sqrt{3}} = \frac{3(3) - 8}{8\sqrt{3}} = \frac{9 - 8}{8\sqrt{3}} = \frac{1}{8\sqrt{3}} \]
Denominator:
\[ \cos^2 45^{\circ} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Now, evaluate the full expression:
\[ \text{Value} = \frac{\frac{1}{8\sqrt{3}}}{\frac{1}{2}} = \frac{1}{8\sqrt{3}} \times 2 = \frac{1}{4\sqrt{3}} \]
Rationalizing the denominator:
\[ = \frac{1}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{4 \times 3} = \frac{\sqrt{3}}{12} \]
Step 4: Final Answer:
The evaluated value is \(\frac{\sqrt{3}}{12}\).