Problem 55 - Algebra and Trigonometry

Question

Use the given information to find the exact value of each of the following: a. \(\sin \frac{\alpha}{2}\) b. \(\cos \frac{\alpha}{2}\) c. \(\tan \frac{\alpha}{2}\) $$ \tan \alpha=\frac{4}{3}, 180^{\circ}<\alpha<270^{\circ} $$

Step-by-Step Solution

Answer

The exact value for \(\sin \frac{\alpha}{2}\) is \(\sqrt{2}\), for \(\cos \frac{\alpha}{2}\) is \(-\frac{\sqrt{6}}{2}\) and for \(\tan \frac{\alpha}{2}\) is \(-\sqrt{2}\).

Step 1: Determine the Sign of the Half Angles

Since \(180^{\circ}<\alpha<270^{\circ}\), it's in the third quadrant where both sine and cosine are negative and tangent is positive. As for the half angle \(\frac{\alpha}{2}\), it falls in the second quadrant because \(90^{\circ}< \frac{\alpha}{2} < 135^{\circ}\). This means that sine is positive, cosine is negative, and tangent is negative.

Step 2: Use the Half-Angle Formulas

The half-angle formulas are as follows:\[\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1-\cos \alpha}{2}}\]\[\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1+\cos \alpha}{2}}\]\[\tan \frac{\alpha}{2} = \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}\]Since we only know \(\tan \alpha\) and not \(\cos \alpha\), we'll need to find \(\cos \alpha\) using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\) and the fact that \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Step 3: Find \(\cos \alpha\)

We know \(\tan \alpha = \frac{4}{3}\) and \(\sin^2 \alpha + \cos^2 \alpha = 1\). Let's find \(\cos \alpha\):In the third quadrant, \(\cos \alpha\) is negative. So, the possible value of \(\cos \alpha\) can be found using Pythagorean theorem \(-\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - (\tan^2 \alpha /(1 + \tan^2 \alpha))} = -\sqrt{1 - (4/3)^2/(1 + (4/3)^2)} = -\frac{3}{5}\)

Step 4: Substitute \(\cos \alpha\) into the Half-Angle Formulas

Substitute \(\cos \alpha = -\frac{3}{5}\) into the half-angle formulas respectively, remembering to take the right signs for the second quadrant (i.e. positive for sine, negative for cosine and negative for tangent). We get:\[\sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-3/5)}{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}\]\[\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-3/5)}{2}} = -\frac{\sqrt{6}}{2}\]\[\tan \frac{\alpha}{2} = -\sqrt{\frac{1 - (-3/5)}{1 + (-3/5)}} = -\sqrt{\frac{4}{2}} = -\sqrt{2}\]

Key Concepts

Sine Half-Angle Formula

The sine half-angle formula, \(\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1-\cos \alpha}{2}}\), is a powerful tool in trigonometry for finding the sine of half of a given angle, \(\alpha\). The challenge with this formula is determining the correct sign before the square root. The sign is based on the quadrant in which \(\frac{\alpha}{2}\) lies.

Considering the original problem, where \(\tan(\alpha) = \frac{4}{3}\) and \(\alpha\) is in the third quadrant (implying a negative cosine), we can then determine the sine of half-angle. Since half of the angle lands in the second quadrant — where sine is positive — we drop the negative sign for the sine half-angle, leading to the positive square root in the solution. To apply this knowledge, first ensure the determination of \(\cos(\alpha)\), and then substitute into the sine half-angle formula to find \(\sin(\frac{\alpha}{2})\).

Cosine Half-Angle Formula

The formula for the cosine of a half angle \(\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1+\cos \alpha}{2}}\) plays an equally significant role in trigonometry as its sine counterpart. As with the sine, it is crucial to ascertain the sign based on where the half angle lands on the unit circle.

In our example, the third quadrant where \(\cos(\alpha)\) is negative splits into the second and fourth quadrants when halved. Since \(\frac{\alpha}{2}\) is in the second quadrant, the cosine value should be negative due to cosine's inherent negativity in that quadrant. With the negative sign clarified, you substitute the previously found \(\cos(\alpha)\) into the formula. The result will give you the value of \(\cos(\frac{\alpha}{2})\).

Tangent Half-Angle Formula

The tangent of a half-angle can be found using the formula \(\tan \frac{\alpha}{2} = \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}\), which is derived from the sine and cosine half-angle formulas. Determining the sign is again essential and depends on which quadrant the half-angle occupies.

For an angle in the third quadrant, as in the original problem, the tangent half-angle formula will result in a negative value since \(\frac{\alpha}{2}\) falls in the second quadrant where the tangent is negative. Remember to use the already computed \(\cos(\alpha)\) within the formula to solve for \(\tan(\frac{\alpha}{2})\). This approach smoothly translates the knowledge of cosine to find the tangent of the half-angle.

Pythagorean Identity

The Pythagorean identity in trigonometry, which states that \(\sin^2 \alpha + \cos^2 \alpha = 1\), is foundational. This identity always holds true for any angle \(\alpha\) and provides a method to find one trigonometric function value given another.

In our exercise, using the given value of \(\tan(\alpha)\), we can exploit the identity to find \(\cos(\alpha)\), understanding that \(\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}\). This process involves a bit of algebra including squaring \(\tan(\alpha)\) and rearranging the terms to solve for \(\cos(\alpha)\) under the restriction of the quadrant, which dictates the sign of the cosine. Once determined, \(\cos(\alpha)\) becomes a bridge to finding the half-angle values for sine, cosine, and tangent.