Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 305: 62

Answer

The product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: $sin(A+B) = sin~A~cos~B+sin~B~cos~A$ Note that this is the formula for the sine of the sum of two angles.

Work Step by Step

The line through the middle of the circle is the diameter. We know that any triangle constructed with the diameter as one side and with a point on the circle's circumference is a right triangle. Therefore, the four sides around the outside have lengths $sin~A, cos~A, sin~B,$ and $cos~B$ since the hypotenuse has a length of 1. In Exercise 53, we saw that for any triangle inscribed in a circle, $\frac{d}{sin~D} = 2r$. In this case, let $d$ be the side opposite the angle $A+B$. Then: $\frac{d}{sin~(A+B)} = 2r = 1$ $d = sin~(A+B)$ Therefore, the side opposite the angle $A+B$ has a length of $sin(A+B)$ The product of the diagonals is $sin~(A+B)(1) = sin(A+B)$ The product of one pair of opposite sides is $sin~A~cos~B$ The product of the other pair of opposite sides is $sin~B~cos~A$ The sum of these two products is $sin~A~cos~B+sin~B~cos~A$ From the information in the question, the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore: $sin(A+B) = sin~A~cos~B+sin~B~cos~A$ Note that this is the formula for the sine of the sum of two angles.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.