Problem 36 - Precalculus : Building Concepts and Connections

Question

Evaluate each trigonometric function without the use of a calculator. $$\cos \left(\arcsin \left(-\frac{3}{5}\right)\right)$$

Step-by-Step Solution

Answer

\(-\frac{4}{5}\)

Step 1: Simplify the expression

Change the expression in terms of sine function using the expression \( \cos(x) = \sqrt{1 - \sin^2(x)} \). As the sine function is negative (\(-3/5\)), the cosine function will fall in either 2nd or 3rd quadrant where \(\cos(x)\) is negative. Therefore, \(\cos(x) = -\sqrt{1 -\sin^2(x)}\). This simplification is based on the Pythagorean trigonometric identity which states that for any angle \(x\), the sum of the squares of \(\sin(x)\) and \(\cos(x)\) is 1.

Step 2: Substitute the value of \(\sin(x)\)

Substitute the value of \(\sin(x)\) as \(-3/5\). Hence, the equation becomes \(\cos(x) = -\sqrt{1 - (- \frac{3}{5})^2}\).

Step 3: Evaluate the expression

Apply the process of calculating the square and square root to simplify the expression. That gives us :- \(\cos(x) = -\sqrt{1 - (\frac{9}{25})}\) = \(-\sqrt{\frac{16}{25}}\) = \(-\frac{4}{5}\)

Key Concepts

Pythagorean Trigonometric Identity

The Pythagorean trigonometric identity is a fundamental relation in trigonometry that relates the sine and cosine of an angle. This identity is derived from the Pythagorean theorem and can be expressed as:
\[\begin{equation} \[sin^2(x) + cos^2(x) = 1\]\end{equation}\]
This equation is true for any angle and helps to find the value of one trigonometric function given another. For instance, if you're given the sine of an angle and need to find the cosine, you can rearrange the identity to:
\[\begin{equation} \[cos(x) = \pm\sqrt{1 - sin^2(x)}\]\end{equation}\]
The \(\pm\) sign indicates that the cosine function could be either positive or negative, depending on the angle's quadrant. This identity is crucial for evaluating trigonometric functions when a calculator is not available, as it serves as a bridge between different functions.

Arcsine Function

The arcsine function, denoted as \(\arcsin\), is the inverse operation of the sine function. It takes a number (representing the sine of an angle) as input and returns the angle itself, where the angle is in the range from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) or \(-90^\circ\) to \(90^\circ\).
The arcsine function can only take values between \(-1\) and \(1\), since these are the minimum and maximum values of the sine function. When given a negative input, like \(-\frac{3}{5}\) in our exercise, the arcsine function tells us that the resulting angle originates from either the third or fourth quadrants where sine values are negative. However, by definition, the range of the arcsine function limits the angle to be in the range where sine is positive (first quadrant) or negative (fourth quadrant). Therefore, in this exercise, the angle from the arcsine function would be in the fourth quadrant.

Quadrant Analysis

Quadrant analysis is a method used in trigonometry to determine the sign of trigonometric functions based on the angle's quadrant. A full rotation about the origin of a coordinate system is divided into four quadrants, each spanning 90 degrees or \(\frac{\pi}{2}\) radians.

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

The function's sign impacts the function value, which can be critical for correctly evaluating an expression without a calculator. In our case, since the sine value \(-\frac{3}{5}\) indicates a negative sine function, we check the second and third quadrants where sine values are negative. As explained in the arcsine function section, the angle's range restricts us to the fourth quadrant, where the cosine is positive. However, since we were given the angle through the arcsine function, which gave us a negative value, we know we must adjust our original quadrant presumption to match the inherent range of the arcsine. Consequently, we find ourselves in the unique situation where the cosine value is negative despite being in the fourth quadrant – a contradiction we resolve by referencing the specific properties of the arcsine function.