Problem 18 - Algebra 2 and Trigonometry

Question

a. Find the exact value of \(\sin 15^{\circ}\) by using \(\sin \left(45^{\circ}-30^{\circ}\right)\) b. Use the value of \(\sin 15^{\circ}\) found in a to find \(\sin 165^{\circ}\) by using \(\sin \left(180^{\circ}-15^{\circ}\right)\) c. Use the value of \(\sin 15^{\circ}\) found in a to find \(\sin 345^{\circ}\) by using \(\sin \left(360^{\circ}-15^{\circ}\right)\) . d. Use the value of \(\sin 15^{\circ}\) found in a to find \(\sin 195^{\circ}\) by using \(\sin \left(180^{\circ}+15^{\circ}\right)\)

Step-by-Step Solution

Answer

a) \(\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}\). b) \(\sin 165^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}\). c) \(\sin 345^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4}\). d) \(\sin 195^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4}\).

Step 1: Using the Difference Identity for Sin

To find \( \sin 15^{\circ} \), we will use the identity \( \sin(a-b) = \sin a \cos b - \cos a \sin b \). Here, \( a = 45^{\circ} \) and \( b = 30^{\circ} \). Thus, we have: \[ \sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \] Substitute the known values: \[ \sin 45^{\circ} = \frac{\sqrt{2}}{2}, \quad \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 45^{\circ} = \frac{\sqrt{2}}{2}, \quad \text{and} \quad \sin 30^{\circ} = \frac{1}{2} \] \[ \sin 15^{\circ} = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \] Simplifying this gives: \[ \sin 15^{\circ} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \]

Step 2: Applying the Supplement Angle Identity

To find \( \sin 165^{\circ} \), use the identity \( \sin(180^{\circ} - \theta) = \sin \theta \). Here \( \theta = 15^{\circ} \), so: \[ \sin 165^{\circ} = \sin(180^{\circ} - 15^{\circ}) = \sin 15^{\circ} \] Thus, \( \sin 165^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4} \).

Step 3: Using the Reference Angle Identity

To find \( \sin 345^{\circ} \), use the identity \( \sin(360^{\circ} - \theta) = -\sin \theta \). Again, \( \theta = 15^{\circ} \), so: \[ \sin 345^{\circ} = \sin(360^{\circ} - 15^{\circ}) = -\sin 15^{\circ} \] Therefore, \( \sin 345^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4} \).

Step 4: Applying the Supplement Identity for Angles Greater Than 180°

To find \( \sin 195^{\circ} \), apply \( \sin(180^{\circ} + \theta) = -\sin \theta \). Here \( \theta = 15^{\circ} \): \[ \sin 195^{\circ} = \sin(180^{\circ} + 15^{\circ}) = -\sin 15^{\circ} \] So, \( \sin 195^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4} \).

Key Concepts

Sine Function

The sine function, often denoted as \( \sin \theta \), is an essential trigonometric function. It relates to the y-coordinate of a point on the unit circle as well as the opposite side's length over the hypotenuse in a right triangle.
The sine function follows specific rules and identities that help in calculating angles and solving various mathematical problems. It oscillates between -1 and 1, with a periodic increment of \( 2\pi \). This function helps in defining relationships involving angles in different parts of geometry and physics, such as waveforms and rotational dynamics.

• At \( 0^{\circ} \) or \( 0 \) radians, \( \sin 0 = 0 \).
• At \( 90^{\circ} \) or \( \pi/2 \) radians, \( \sin \pi/2 = 1 \).
• The sine function is symmetric about the origin, meaning \( \sin(-\theta) = -\sin \theta \).

Angle Sum and Difference Formulas

The Angle Sum and Difference Formulas allow the calculation of the sine, cosine, and tangent of angles that are sums or differences of known angles.
Using the formula for the sine of a difference, as seen in the problem of finding \( \sin 15^{\circ} \), we use: \[ \sin(a-b) = \sin a \cos b - \cos a \sin b \] for specific angles \( a \) and \( b \), this formula makes complex calculations more manageable.
For example:

• \( \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \)
• This expands into: \( \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \)
• Resulting in: \( \frac{\sqrt{6} - \sqrt{2}}{4} \)

This formula is invaluable for obtaining the sine of non-standard angles without a calculator.

Supplementary Angles

Supplementary angles are pairs of angles whose sum totals \( 180^{\circ} \). In trigonometry, these angles have a neat identity: the sine of an angle and its supplementary angle are equal in magnitude. This is expressed as \( \sin(180^{\circ} - \theta) = \sin \theta \).
In the exercise, this identity aids in quickly determining \( \sin 165^{\circ} \) using \( \sin 15^{\circ} \). Geometrically, this represents the reflection of a point in the first quadrant across the y-axis.

• Example: \( \sin 165^{\circ} = \sin(180^{\circ} - 15^{\circ}) = \sin 15^{\circ} \).

The angle sum using this identity ensures the calculated value remains within the unit circle's context, confirming its position and value accurately.

Reference Angles

Reference angles are used in trigonometry to find the value of trigonometric functions for larger angles by relating them back to familiar angles under \( 90^{\circ} \). These angles, alongside identities like \( \sin(360^{\circ} - \theta) = -\sin \theta \), allow the simplifying of calculations for angles beyond a single revolution.
In practice, the angle \( 345^{\circ} \) is more straightforward by considering it in conjunction with the reference angle: \( 15^{\circ} \). This is seen through the expression:

• \( \sin 345^{\circ} = -\sin 15^{\circ} \), using the identity for angles exceeding \( 360^{\circ} \).

This technique provides consistency in assessing trigonometric functions, emphasizing symmetries in periodic cycles and transformations on the unit circle.