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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
Chapter 7, Problem 33

Evaluate each expression without using a calculator.
tan (arcsin (3/5) + arccos (5/7))

Verified step by step guidance
1
Identify the angles involved: let \( \alpha = \arcsin\left(\frac{3}{5}\right) \) and \( \beta = \arccos\left(\frac{5}{7}\right) \). We want to find \( \tan(\alpha + \beta) \).
Recall the tangent addition formula: \[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]. So, we need to find \( \tan \alpha \) and \( \tan \beta \).
From \( \alpha = \arcsin\left(\frac{3}{5}\right) \), use the definition of sine to find the opposite side as 3 and hypotenuse as 5. Use the Pythagorean theorem to find the adjacent side: \[ \text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} \]. Then, calculate \( \tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{\sqrt{16}} \).
From \( \beta = \arccos\left(\frac{5}{7}\right) \), use the definition of cosine to find the adjacent side as 5 and hypotenuse as 7. Use the Pythagorean theorem to find the opposite side: \[ \text{opposite} = \sqrt{7^2 - 5^2} = \sqrt{49 - 25} \]. Then, calculate \( \tan \beta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{24}}{5} \).
Substitute \( \tan \alpha \) and \( \tan \beta \) into the tangent addition formula and simplify the expression to find \( \tan(\alpha + \beta) \) without using a calculator.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Trigonometric Functions

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