Problem 20 - Algebra 2 and Trigonometry

Question

a. Find the exact value of \(\sin 210^{\circ}\) by using \(\sin \left(270^{\circ}-60^{\circ}\right)\) b. Find the exact value of \(\cos 210^{\circ}\) by using \(\cos \left(270^{\circ}-60^{\circ}\right)\) c. Find the exact value of \(\sin 165^{\circ}\) by using \(\sin \left(210^{\circ}-45^{\circ}\right)\) d. Use the value of \(\sin 165^{\circ}\) found in \(c\) to find \(\sin \left(-15^{\circ}\right)\) by using \(\sin \left(165^{\circ}-180^{\circ}\right) .\) e. Use the value of \(\sin \left(-15^{\circ}\right)\) found in \(d\) to find \(\sin 195^{\circ}\) by using \(\sin \left(180^{\circ}-\left(-15^{\circ}\right)\right)\) I. Use the value of \(\sin \left(-15^{\circ}\right)\) found in \(\mathrm{d}\) to find the exact value of \(\sin 105^{\circ} .\)

Step-by-Step Solution

Answer

a. \(\sin 210^{\circ} = -\frac{1}{2}\); b. \(\cos 210^{\circ} = -\frac{\sqrt{3}}{2}\); c. \(\sin 165^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}\); d. \(\sin(-15^{\circ}) = -\frac{\sqrt{6} - \sqrt{2}}{4}\); e. \(\sin 195^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}\); I. \(\sin 105^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}\).

Step 1: Calculate \( \sin 210^{\circ} \)

Using the identity \( \sin(270^{\circ} - 60^{\circ}) = \cos 60^{\circ} \), evaluate \( \cos 60^{\circ} \), which equals \( \frac{1}{2} \). The sine of an angle in the third quadrant is negative, so \( \sin 210^{\circ} = -\frac{1}{2}.\)

Step 2: Calculate \( \cos 210^{\circ} \)

Apply the identity \( \cos(270^{\circ} - 60^{\circ}) = -\sin 60^{\circ} \). Calculate \( \sin 60^{\circ} \), resulting in \( \frac{\sqrt{3}}{2} \). Therefore, \( \cos 210^{\circ} = -\frac{\sqrt{3}}{2}.\)

Step 3: Find \( \sin 165^{\circ} \)

Use the identity \( \sin(210^{\circ} - 45^{\circ}) = \sin 210^{\circ}\cos 45^{\circ} - \cos 210^{\circ}\sin 45^{\circ} \). Substitute \( \sin 210^{\circ} = -\frac{1}{2} \), \( \cos 210^{\circ} = -\frac{\sqrt{3}}{2} \), with \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \), giving \( \sin 165^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)

Step 4: Calculate \( \sin (-15^{\circ}) \) using \( \sin 165^{\circ} \)

Find \( \sin(-15^{\circ}) = \sin (165^{\circ} - 180^{\circ}) \), which equals \( -\sin 165^{\circ} \). From Step 3, \( \sin 165^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4} \). Therefore, \( \sin (-15^{\circ}) = -\frac{\sqrt{6} - \sqrt{2}}{4}.\)

Step 5: Calculate \( \sin 195^{\circ} \)

Use \( \sin (180^{\circ} - (-15^{\circ})) = \sin 195^{\circ} \). The sine of \((180^{\circ} + x)\) is \( -\sin x \) when \(x\) is in the third quadrant. Since \( \sin (-15^{\circ}) = -\frac{\sqrt{6} - \sqrt{2}}{4} \), \( \sin 195^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)

Step 6: Find \( \sin 105^{\circ} \) using \( \sin (-15^{\circ}) \)

Use the sum identity \( \sin(120^{\circ} - 15^{\circ}) = \sin 105^{\circ} \). Substitute \( \sin 120^{\circ} = \frac{\sqrt{3}}{2} \) and \( \cos 120^{\circ} = -\frac{1}{2} \), with \( \sin (-15^{\circ}) = -\frac{\sqrt{6} - \sqrt{2}}{4} \). Calculate \( \cos 15^{\circ} \) using the identity: \( \cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \). Hence, \( \sin 105^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}.\)

Key Concepts

Sine and Cosine Values

The values of sine and cosine for specific angles are foundational for solving trigonometric problems. These trigonometric functions give us the ratios of the sides of a right triangle for a given angle.
For example:

• Sine ( \( \sin \)) of an angle is the ratio of the length of the opposite side to the hypotenuse.
• Cosine ( \( \cos \)) is the ratio of the length of the adjacent side to the hypotenuse.

Knowing the sine and cosine values of standard angles like \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\) can help solve trigonometric equations effortlessly:

• \( \sin 30^{\circ} = \frac{1}{2} \)
• \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
• \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
• \( \cos 60^{\circ} = \frac{1}{2} \)

Angle Subtraction Identity

The angle subtraction identity is a powerful tool in trigonometry that allows us to find the sine or cosine of a non-standard angle by breaking it down into the difference of two common angles.
The identities are:

• \( \sin(a - b) = \sin a \cos b - \cos a \sin b \)
• \( \cos(a - b) = \cos a \cos b + \sin a \sin b \)

These formulas are handy when you need to calculate values for angles that aren’t commonly memorized, such as \(210^{\circ}\) or \(165^{\circ}\).

In the exercise, we used the identities to find the sine and cosine values of various angles by expanding them into expressions using common angles such as \(270^{\circ}\) and \(60^{\circ}\), making problem-solving simpler and more manageable.

Quadrant Sign Rules

Understanding the sign of trigonometric functions in different quadrants is crucial for getting the correct values.
The unit circle is divided into four quadrants, and each quadrant impacts the sign (+ or -) of sine, cosine, and tangent values:

• **First Quadrant**: All trigonometric functions are positive.
• **Second Quadrant**: Sine is positive, while cosine and tangent are negative.
• **Third Quadrant**: Tangent is positive, but sine and cosine are negative.
• **Fourth Quadrant**: Cosine is positive, but sine and tangent are negative.

In our exercise, to find values like \( \sin 210^{\circ} \) and \( \cos 210^{\circ} \), recognizing that these angles fall into the third quadrant was key. Hence, we adjusted our values accordingly to ensure we applied the correct signs, which guides accurate trigonometric calculation.