Evaluate the expressions without using a calculator. \( 4 \tan 330^{\circ} \cos 135^{\circ} \sin 300^{\circ} \)

First, let's evaluate each trigonometric function in the expression. 1. **Finding \( \tan 330^{\circ} \)**: - Since \( 330^{\circ} = 360^{\circ} - 30^{\circ} \), it lies in the fourth quadrant where tangent is negative. - Thus, \( \tan 330^{\circ} = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}} \). 2. **Finding \( \cos 135^{\circ} \)**: - \( 135^{\circ} = 180^{\circ} - 45^{\circ} \) is in the second quadrant where cosine is negative. - Therefore, \( \cos 135^{\circ} = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}} \). 3. **Finding \( \sin 300^{\circ} \)**: - \( 300^{\circ} = 360^{\circ} - 60^{\circ} \) is in the fourth quadrant where sine is negative. - Therefore, \( \sin 300^{\circ} = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \). Now, substituting these values into the expression \( 4 \tan 330^{\circ} \cos 135^{\circ} \sin 300^{\circ} \): \[ 4 \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{1}{\sqrt{2}}\right) \left(-\frac{\sqrt{3}}{2}\right) \] Calculating step-by-step: 1. \( -\frac{1}{\sqrt{3}} \times -\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{3}\sqrt{2}} = \frac{1}{\sqrt{6}} \) 2. However, now we multiply that result with \( -\frac{\sqrt{3}}{2} \): \[ \frac{1}{\sqrt{6}} \times -\frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2\sqrt{6}} \] 3. Now, we can simplify this: \(-\frac{\sqrt{3}}{2\sqrt{6}} = -\frac{\sqrt{3}}{2 \cdot \sqrt{6}} = -\frac{\sqrt{3}}{2 \cdot \sqrt{2 \cdot 3}} = -\frac{\sqrt{3}}{2 \cdot \sqrt{2}\cdot \sqrt{3}} = -\frac{1}{2\sqrt{2}} \). 4. Finally, multiply it by 4: \[ 4 \times -\frac{1}{2\sqrt{2}} = -\frac{4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} \] Thus, the value of the expression \( 4 \tan 330^{\circ} \cos 135^{\circ} \sin 300^{\circ} \) is: \[ \boxed{-\sqrt{2}} \]