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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 43

In Exercises 29–51, find the exact value of each expression. Do not use a calculator. _ csc(tan⁻¹ √3/3)

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Recognize that the expression is \( \csc(\tan^{-1}(\frac{\sqrt{3}}{3})) \). The first step is to let \( \theta = \tan^{-1}(\frac{\sqrt{3}}{3}) \), which means \( \tan(\theta) = \frac{\sqrt{3}}{3} \).
Recall that \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Assign the opposite side as \( \sqrt{3} \) and the adjacent side as 3 to form a right triangle representing \( \theta \).
Use the Pythagorean theorem to find the hypotenuse \( h \) of the triangle: \( h = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} \). Simplify \( \sqrt{12} \) to \( 2\sqrt{3} \).
Since \( \csc(\theta) = \frac{1}{\sin(\theta)} \) and \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), calculate \( \sin(\theta) = \frac{\sqrt{3}}{2\sqrt{3}} \).
Finally, express \( \csc(\theta) \) as the reciprocal of \( \sin(\theta) \), which is \( \csc(\theta) = \frac{1}{\sin(\theta)} \). Simplify this expression to find the exact value.

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Inverse Trigonometric Functions

Inverse trigonometric functions, like tan⁻¹(x), return the angle whose trigonometric ratio equals x. For example, tan⁻¹(√3/3) gives the angle whose tangent is √3/3, which helps in finding the angle measure needed to evaluate other trigonometric functions.
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Right Triangle Trigonometry

Right triangle trigonometry relates the sides of a right triangle to its angles using ratios like sine, cosine, and tangent. Knowing the tangent ratio allows you to determine the opposite and adjacent sides, which can then be used to find other ratios such as cosecant.
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Reciprocal Trigonometric Functions

Cosecant (csc) is the reciprocal of sine, defined as csc(θ) = 1/sin(θ). Once the angle θ is found, calculating csc(θ) involves finding the sine of θ and then taking its reciprocal to get the exact value.
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