Problem 15 - Algebra 2 and Trigonometry

Question

In \(3-17,\) find the exact value of \(\cos (A-B)\) for each given pair of values. \(A=\pi, B=\frac{2 \pi}{3}\)

Step-by-Step Solution

Answer

The exact value of \(\cos(A-B)\) is \(\frac{1}{2}\).

Step 1: Recall the Cosine Subtraction Formula

The formula for finding the cosine of the difference of two angles is: \[ \cos(A-B) = \cos A \cos B + \sin A \sin B. \] This formula will help us find the value of \( \cos(A-B) \) given \( A \) and \( B \).

Step 2: Calculate \( \cos A \) and \( \sin A \)

Substitute \( A = \pi \) into the trigonometric functions:\[ \cos(\pi) = -1 \] \[ \sin(\pi) = 0 \] These are the standard trigonometric values for an angle of \( \pi \) radians.

Step 3: Calculate \( \cos B \) and \( \sin B \)

Substitute \( B = \frac{2\pi}{3} \) into the trigonometric functions:\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \] \[ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \] These values come from the unit circle for an angle in the second quadrant.

Step 4: Substitute into the Cosine Subtraction Formula

Now substitute these values into the formula:\[ \cos(A-B) = \cos(\pi) \cos\left(\frac{2\pi}{3}\right) + \sin(\pi) \sin\left(\frac{2\pi}{3}\right). \]

Step 5: Simplify the Expression

Compute each term:- \( \cos(\pi) \cos\left(\frac{2\pi}{3}\right) = (-1) \times \left(-\frac{1}{2}\right) = \frac{1}{2} \)- \( \sin(\pi) \sin\left(\frac{2\pi}{3}\right) = 0 \times \frac{\sqrt{3}}{2} = 0 \)So the expression simplifies to:\[ \cos(A-B) = \frac{1}{2} + 0 = \frac{1}{2}. \]

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in understanding relationships within triangles and circular motion. They include sine, cosine, and tangent. Each function represents a different relationship between the angles and sides of a right-angled triangle or points on a circle. In this exercise, we specifically use cosine and sine:

• Cosine (ccose): Relates the adjacent side to the hypotenuse in a triangle.
• Sine (csine): Relates the opposite side to the hypotenuse.

For any angle, these functions can be described using the unit circle, where the angle's position determines the values of sine and cosine. They have specific values for remarkable angles like 0, cfrac{cpie}{2}e, cpie, and so forth, used frequently in trigonometry.
Understanding these basic functions helps in computing expressions like ccos(A-B)e, as they dictate the relationships being multiplied in terms like ccos A eccos Be and csin A ecsin Be.

Unit Circle

The unit circle is a vital concept in trigonometry, providing a visual tool to understand angles and trigonometric functions. It is a circle with a radius of 1, centered at the origin of a coordinate system. Each point on the circle represents an angle cthetae from the positive x-axis:

• The x-coordinate reflects ccos cthetae.
• The y-coordinate reflects csin cthetae.

For cA = cpie, the point on the unit circle is (-1, 0). Thus, ccos(cpie) = -1e and csin(cpie) = 0e.
For cB = cfrac{2cpie}{3}e, in the second quadrant, the values are ccos(cfrac{2cpie}{3}e) = -cfrac{1}{2}ee and csin(cfrac{2cpie}{3}e) = cfrac{csqrt{3}e}{2}ee.
These positions on the unit circle facilitate straightforward lookup of trigonometric values necessary to calculate ccos(A-B)e.

Angle Subtraction

Angle subtraction is a technique used to find the trigonometric function of the difference between two angles. For cosine, it uses the cosine subtraction formula:

• ccos(A-B) = ccos A eccos Be + csin A ecsin Be

This formula allows us to find the exact trigonometric value by utilizing known values of cAe and cBe.
In this exercise:

• ccos(cpie) = -1e and csin(cpie) = 0e
• ccos(cfrac{2cpie}{3}e) = -cfrac{1}{2}ee and csin(cfrac{2cpie}{3}e) = cfrac{csqrt{3}e}{2}ee

By substituting these into the formula, we find:

• The cosine term: c(-1)e c(-cfrac{1}{2}e)e = cfrac{1}{2}e
• The sine term: 0 c(cfrac{csqrt{3}e}{2}e)e = 0

Thus, ccos(A-B) = cfrac{1}{2}ee simplifies our result efficiently.