Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left(\tan ^{-1} \sqrt{3}\right)\) (b) \(\cos \left(\sin ^{-1} 1\right)\) (c) \(\tan \left(\cos ^{-1} 0\right)\)

Step-by-Step Solution

Answer

(a) \( \frac{\sqrt{3}}{2} \); (b) 0; (c) Undefined

Step 1: Understand the First Expression

We need to find \( \sin(\tan^{-1}(\sqrt{3})) \). This involves using the identity for tangent and sine in a right triangle.

Step 2: Find \( \tan^{-1}(\sqrt{3}) \)

Let \( \theta = \tan^{-1}(\sqrt{3}) \). Then \( \tan(\theta) = \sqrt{3} \). This implies we have a right triangle where the opposite side is \(\sqrt{3}\) and the adjacent side is \(1\).

Step 3: Calculate Hypotenuse in the Triangle

Using Pythagorean theorem, the hypotenuse \(h\) is \( \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \).

Step 4: Find \( \sin(\theta) \)

In the triangle from Steps 2 and 3, \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \). Therefore, \( \sin(\tan^{-1}(\sqrt{3})) = \frac{\sqrt{3}}{2} \).

Step 5: Understand the Second Expression

We need to find \( \cos(\sin^{-1}(1)) \). Recognize that \( \sin^{-1}(1) = \frac{\pi}{2} \) since \( \sin\left(\frac{\pi}{2}\right) = 1 \).

Step 6: Evaluate \( \cos(\frac{\pi}{2}) \)

Knowing the cosine of \( \frac{\pi}{2} \) is \( 0 \), we can say \( \cos(\sin^{-1}(1)) = 0 \).

Step 7: Understand the Third Expression

We need to find \( \tan(\cos^{-1}(0)) \). Recognize that \( \cos^{-1}(0) = \frac{\pi}{2} \) since \( \cos\left(\frac{\pi}{2}\right) = 0 \).

Step 8: Evaluate \( \tan(\frac{\pi}{2}) \)

The tangent function is undefined for \( \frac{\pi}{2} \) because it involves division by \(0\). Thus, \( \tan(\cos^{-1}(0)) \) is undefined.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle associated with a given trigonometric ratio. For example, if we know the tangent of an angle is \( \sqrt{3} \), we can use the inverse tangent function, denoted as \( \tan^{-1}(\sqrt{3}) \), to find the angle itself.

• \( \tan^{-1}(\sqrt{3}) \) gives the angle whose tangent is \( \sqrt{3} \).
• Other inverse functions include \( \sin^{-1} \) and \( \cos^{-1} \), which help find angles for their respective trigonometric values.

Understanding these functions is essential for solving problems that require finding exact angle measures based on trigonometric values.
The angle values resulting from inverse trigonometric functions often correspond to specific angles on the unit circle or within triangles.

Right Triangle

A right triangle is a triangle that includes a 90-degree angle. This makes it special because it allows us to use various trigonometric properties and theorems to find unknown sides or angles.

• The sides of a right triangle are categorized into the hypotenuse, opposite, and adjacent sides relative to the non-right angles.
• Right triangles form the basis for defining sine, cosine, and tangent functions using the ratios of these sides.
• In context, if \( \theta = \tan^{-1}(\sqrt{3}) \), then this angle is part of a right triangle with sides that relate to the tangent function.
• The opposite side is \( \sqrt{3} \), the adjacent side is \( 1 \), and you can use these to calculate the sine or cosine of the angle.

This property allows trigonometric ratios, like sine or tangent, to be directly applied by considering a triangle's side lengths.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in mathematics that applies to right triangles. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
To find the third side of a right triangle when two sides are known, simply apply the formula:
\[ c^2 = a^2 + b^2 \]

• For example, if the two sides are \( \sqrt{3} \) and \( 1 \), as in our problem, calculate the hypotenuse \( h \) by substituting into the theorem: \( h^2 = (\sqrt{3})^2 + 1^2 \).
• This simplifies to: \( h^2 = 3 + 1 = 4 \).
• Taking the square root gives us the hypotenuse \( h = 2 \).

Thus, with the hypotenuse known, you can find other trigonometric values like sine and cosine, which require the length of the hypotenuse.