Problem 38 - Precalculus : Building Concepts and Connections

Question

Evaluate each trigonometric function without the use of a calculator. $$\sin \left(\arccos \left(-\frac{4}{5}\right)\right)$$

Step-by-Step Solution

Answer

\(\sin \left(\arccos \left(-\frac{4}{5}\right)\right) = \frac{3}{5}\)

Step 1: Identify the given information

The problem requires to find the value of \(\sin \theta\) where \(\cos \theta = -\frac{4}{5}\). Based on the Pythagorean identity, \(\sin^2 \theta = 1 - \cos^2 \theta\).

Step 2: Use Pythagorean identity and the given value of cos θ to find sin θ

Since \(\cos \theta = -\frac{4}{5}\), square this value and subtract it from 1, \(1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}\).Take the square root of this value to find the absolute value of \(\sin \theta\), \(\sqrt{\frac{9}{25}} = \frac{3}{5}\).

Step 3: Determine the sign of sin θ

As the angle is in the second quadrant (since \(\cos \theta\) is negative and \(\arccos\) returns values from 0 to π), \(\sin \theta\) will be positive. Therefore, \(\sin \theta = \frac{3}{5}\).

Key Concepts

Pythagorean Identity

Understanding the Pythagorean identity is crucial when evaluating trigonometric functions without a calculator. It expresses a fundamental relationship between the sine and cosine of an angle: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \.\] Essentially, this formula tells us that the square of the sine of an angle plus the square of the cosine of that same angle equals one, which stems from the Pythagorean theorem relating the sides of a right triangle.
To apply this to our exercise, once we know the cosine of an angle, we can find the sine of the angle by rearranging the Pythagorean identity: \[ \sin^2(\theta) = 1 - \cos^2(\theta) \.\] This maneuver is particularly helpful as it avoids the need for memorizing sine values for different angles; instead, you only need to remember this simple relationship.

Arccosine Function

The arccosine function, denoted as \( \arccos \), is the inverse operation of the cosine function. In other words, if you have the cosine of an angle, \( \arccos \) helps you find the angle itself. The output of the arccosine function is an angle, typically measured in radians, and it ranges from 0 to \( \pi \) in the case of real numbers. In our exercise,
\( \arccos\left(-\frac{4}{5}\right) \) helps us determine the angle whose cosine is \( -\frac{4}{5} \). Since the cosine of an angle in a right triangle represents the adjacent side over the hypotenuse, \( \arccos \) essentially gives us the angle without needing to construct the actual triangle. It is important to remember that since arccosine gives an angle in the first or second quadrant, the sign determined in the next step will establish the correct quadrant the original angle lies in.

Sin Function Calculation Without Calculator

Calculating the sine function without the use of a calculator can be streamlined by utilizing known relationships and identities. Once we have the angle from the arccosine function, as with the exercise \( \sin (\arccos (-\frac{4}{5})) \), we determine the sine value by applying the Pythagorean identity. As explained, this lets us find the sine value directly from the cosine value without plotting graphs or using a calculator.
It's also imperative to take into account the angle's quadrant to apply the correct sign to the sine value, given that sine is positive in the first and second quadrants and negative in the third and fourth quadrants. Considering that \( \arccos \) yields values between 0 and \( \pi \), or the first two quadrants, and in our specific problem the cosine value is negative, we conclude that the angle is in the second quadrant where sine values are positive. This understanding allows us to correctly determine that \( \sin \theta = \frac{3}{5} \) without relying on calculator outputs.