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

We need to evaluate:
\[ \frac{\sin^3 60^\circ - \tan 30^\circ}{\cos^2 45^\circ} \]

Step 1: Use standard trigonometric values
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \] \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \] \[ \cos 45^\circ = \frac{\sqrt{2}}{2} \Rightarrow \cos^2 45^\circ = \frac{1}{2} \]

Step 2: Compute numerator
\[ \sin^3 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{8} \] \[ \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \] \[ \text{Numerator} = \frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{3} \] Take LCM 24:
\[ = \frac{9\sqrt{3}}{24} - \frac{8\sqrt{3}}{24} \] \[ = \frac{\sqrt{3}}{24} \]

Step 3: Divide by denominator
\[ \frac{\frac{\sqrt{3}}{24}}{\frac{1}{2}} = \frac{\sqrt{3}}{24} \times 2 = \frac{\sqrt{3}}{12} \]

Final Answer:
\[ \boxed{\frac{\sqrt{3}}{12}} \]