Question

determine whether the positive or negative square root should be selected.

$$\sin 136.0^{\circ}=\pm \sqrt{\frac{1-\cos 272^{\circ}}{2}}$$

since ( 136.0^{circ} ) is in quadrant the square root should be selected

Explanation:

Step1: Recall the sign of sine function in different quadrants

The sine function \(y = \sin\theta\) has the following sign rules:

• In the first quadrant (\(0^{\circ}<\theta < 90^{\circ}\)), \(\sin\theta>0\).
• In the second quadrant (\(90^{\circ}<\theta < 180^{\circ}\)), \(\sin\theta>0\).
• In the third quadrant (\(180^{\circ}<\theta < 270^{\circ}\)), \(\sin\theta<0\).
• In the fourth quadrant (\(270^{\circ}<\theta < 360^{\circ}\)), \(\sin\theta<0\).

Step2: Determine the quadrant of \(136.0^{\circ}\)

Since \(90^{\circ}<136.0^{\circ}<180^{\circ}\), the angle \(136.0^{\circ}\) is in the second quadrant.

Step3: Determine the sign of \(\sin136.0^{\circ}\)

Because the sine function is positive in the second quadrant, for the formula \(\sin136.0^{\circ}=\pm\sqrt{\frac{1 - \cos272^{\circ}}{2}}\), we should select the positive square - root.

Answer:

Since \(136.0^{\circ}\) is in quadrant II, the positive square root should be selected.