Problem 7 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left[\arcsin \left(-\frac{3}{10}\right)\right]\) (b) \(\cos \left(\arccos \frac{1}{2}\right)\) (c) \(\tan (\arctan 14)\)

Step-by-Step Solution

Answer

(a) \(-\frac{3}{10}\), (b) \(\frac{1}{2}\), (c) 14.

Step 1: Understanding Problem (a)

We need to find \(\sin \left[\arcsin \left(-\frac{3}{10}\right)\right]\). The \(\arcsin\) function outputs an angle \(\theta\) such that \(\sin(\theta) = -\frac{3}{10}\). Thus, \(\sin \left(\theta\right)\) simply returns the original value \(-\frac{3}{10}\).

Step 2: Solve Problem (a)

When the sine function is composed with its inverse, it returns the original value within the range of arcsine. Hence, \(\sin \left(\arcsin \left(-\frac{3}{10}\right)\right) = -\frac{3}{10}\).

Step 3: Understanding Problem (b)

We need to find \(\cos \left(\arccos \frac{1}{2}\right)\). The \(\arccos\) function outputs an angle \(\theta\) such that \(\cos(\theta) = \frac{1}{2}\). Hence, \(\cos(\theta)\) equals \(\frac{1}{2}\).

Step 4: Solve Problem (b)

When the cosine function is composed with its inverse, it returns the original value within the range of arccosine. So, \(\cos \left(\arccos \frac{1}{2}\right) = \frac{1}{2}\).

Step 5: Understanding Problem (c)

We need to find \(\tan(\arctan 14)\). The \(\arctan\) function gives an angle \(\theta\) whose tangent is 14. Thus, \(\tan(\theta) = 14\).

Step 6: Solve Problem (c)

The tangent function composed with its inverse will give back the original value. Therefore, \(\tan(\arctan 14) = 14\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when given a ratio. They essentially reverse the process of trigonometric functions. For every basic trigonometric function (sine, cosine, tangent), there is an inverse.

• The inverse of sine is arcsin (or \(ackslash arcsin\)), which represents the angle whose sine is a given number.
• The inverse of cosine is arccos (or \(ackslash arccos \)), representing the angle whose cosine is a specific value.
• The inverse of tangent is arctan (or \(ackslash arctan\)), representing the angle whose tangent is a certain ratio.

When you compose a trigonometric function with its inverse, as long as you're within the correct range, you'll obtain the original input. Why the range? Because inverse functions are restricted to principal values to ensure they remain functions (each input corresponds to exactly one output). This is important when solving problems like those in our examples, where you're often asked to find exact values using these inverse functions.

Sine Function

The sine function is fundamental in trigonometry. It describes the relationship between the angle of a right triangle and the length of its opposite side relative to the hypotenuse.

• The general form for sine is \(\sin(\theta)\), where \(\theta\) is the angle.
• The sine function has a range from -1 to 1 and a period of \(2\pi\).

In problem (a), we used the property that the arcsin function returns an angle whose sine corresponds to the input value. So, when \(\sin (\arcsin (-\frac{3}{10}))\), it simplifies to \(-\frac{3}{10}\), because the sine of the angle returned by \(\arcsin\) is precisely \(-\frac{3}{10}\). This property helps in quickly solving expressions involving inverse functions without needing additional calculations.

Cosine Function

The cosine function is another cornerstone of trigonometry. Cosine relates the angle of a right triangle to the length of its adjacent side over the hypotenuse.

• The form \(\cos(\theta)\) is used, where \(\theta\) is the angle.
• Its range is the same as sine, from -1 to 1, but it starts at 1 when \(\theta = 0\), giving it a unique shape.

In practice, like in problem (b), using \(\arccos\) simplifies the calculation to retrieve the original cosine value of the angle. Given \(\cos(\arccos \frac{1}{2})\), you receive \(\frac{1}{2}\), illustrating that \(\arccos\) accurately finds the angle whose cosine is \(\frac{1}{2}\), again demonstrating the mirror-like operation of cosine and arccosine.

Tangent Function

The tangent function is another key player, defined quite simply as the ratio of sine to cosine, \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\).

• Its period is \(\pi\), differing from that of sine and cosine.
• Importantly, tangent values can extend beyond the range of sine and cosine, as it is not restricted between -1 and 1.

When tasked with \(\tan(\arctan 14)\) in problem (c), this simplifies directly to 14. The \(\arctan\) function finds an angle whose tangent equals 14, and the \(\tan\) function of this angle reverts back to 14, highlighting the functional symmetry again, even for broader values.