Problem 35 - Algebra and Trigonometry

Question

35-40 Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) from the given information. \(\sin x=\frac{3}{5}, \quad 0^{\circ} < x < 90^{\circ}\)

Step-by-Step Solution

Answer

\( \sin \frac{x}{2} = \sqrt{\frac{1}{10}}, \cos \frac{x}{2} = \sqrt{\frac{9}{10}}, \tan \frac{x}{2} = \frac{1}{3} \).

Step 1: Determine the Value of \( \cos x \)

Given \( \sin x = \frac{3}{5} \) and using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can find \( \cos x \).\Plug \( \sin x \) into the identity: \( \left(\frac{3}{5}\right)^2 + \cos^2 x = 1 \).\Calculate: \( \frac{9}{25} + \cos^2 x = 1 \).\Then, \( \cos^2 x = 1 - \frac{9}{25} = \frac{16}{25} \).\So, \( \cos x = \frac{4}{5} \) (since \( \cos x > 0 \) in the first quadrant).

Step 2: Use Double Angle Identity for \( \sin \frac{x}{2} \)

For \( \sin \frac{x}{2} \), use the identity \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \).\Plug \( \cos x = \frac{4}{5} \) into the formula: \( \sin \frac{x}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{1}{10}} \).\Since \( 0^\circ < x < 90^\circ \), \( \frac{x}{2} \) is also in the first quadrant, so \( \sin \frac{x}{2} > 0 \).\Hence, \( \sin \frac{x}{2} = \sqrt{\frac{1}{10}} \).

Step 3: Use Double Angle Identity for \( \cos \frac{x}{2} \)

For \( \cos \frac{x}{2} \), use the identity \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \).\Plug \( \cos x = \frac{4}{5} \) into the formula: \( \cos \frac{x}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{9}{10}} \).\Since \( \frac{x}{2} \) is in the first quadrant, \( \cos \frac{x}{2} > 0 \).\Therefore, \( \cos \frac{x}{2} = \sqrt{\frac{9}{10}} \).

Step 4: Use Identity for \( \tan \frac{x}{2} \)

For \( \tan \frac{x}{2} \), use the identity \( \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} \).\Substitute \( \sin x = \frac{3}{5} \) and \( \cos x = \frac{4}{5} \) into the formula: \( \tan \frac{x}{2} = \frac{\frac{3}{5}}{1 + \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{9}{5}} \).\Simplify it: \( \tan \frac{x}{2} = \frac{3}{9} = \frac{1}{3} \).\Thus, \( \tan \frac{x}{2} = \frac{1}{3} \).

Key Concepts

Pythagorean identity

The Pythagorean identity is a fundamental equation in trigonometry that relates the squares of the sine and cosine of an angle. It is given by the formula:

• \( \sin^2 x + \cos^2 x = 1 \)

This identity is crucial because it helps link the two primary trigonometric functions, sine and cosine. Given any one of these values, you can easily calculate the other.

In the original exercise, \( \sin x = \frac{3}{5} \) was given, and by applying the Pythagorean identity, you can find the value of \( \cos x \). You'll first square \( \frac{3}{5} \) to get \( \frac{9}{25} \).

• Then, subtract this from 1 because: \( \cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} \)

This calculation results in \( \cos^2 x = \frac{16}{25} \), so \( \cos x = \frac{4}{5} \) since we are in the first quadrant where cosine is positive.

Double angle identities

Double angle identities help compute trigonometric functions for half angles. In this context, we use them to find \( \sin \frac{x}{2} \) and \( \cos \frac{x}{2} \). These identities are crucial when dealing with expressions involving half angles:

• For sine: \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \)
• For cosine: \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \)

Since the angle \( x \) is in the range \( 0^\circ < x < 90^\circ \), the half-angle \( \frac{x}{2} \) is also in the first quadrant, where both sine and cosine are positive.

Plug \( \cos x = \frac{4}{5} \) into the formulas:

• \( \sin \frac{x}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{1}{10}} \)
• \( \cos \frac{x}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{9}{10}} \)

These calculations ensure you have both the sine and cosine values for the half-angle.

First quadrant

The first quadrant in the coordinate plane is crucial for understanding the signs of trigonometric functions. In the first quadrant, where both x and y coordinates are positive, the sine, cosine, and tangent of an angle are also positive.

In trigonometry, the first quadrant corresponds to angles between \( 0^\circ \) and \( 90^\circ \). It is the only quadrant where all primary trigonometric functions assume positive values:

• \( \sin x > 0 \)
• \( \cos x > 0 \)
• \( \tan x > 0 \)

This information is essential for determining the signs when using trigonometric identities. In the exercise provided, knowing that \( x \) and thereby \( \frac{x}{2} \) are within the first quadrant helps confirm the positive nature of the trigonometric values for sine and cosine of half angles.