Trigonometry (11th Edition) Clone

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

Chapter 5 - Test - Page 250: 11


$cos~2A = \frac{cot~A-tan~A}{csc~A~sec~A}$

Work Step by Step

We can verify the identity: $cos~2A = cos^2~A-sin^2~A$ $cos~2A = cos~A~cos~A-sin~A~sin~A$ $cos~2A = (cos~A~cos~A-sin~A~sin~A)(\frac{\frac{1}{sin~A~cos~A}}{\frac{1}{sin~A~cos~A}})$ $cos~2A = \frac{(\frac{cos~A}{sin~A}-\frac{sin~A}{cos~A})}{{\frac{1}{sin~A~cos~A}}}$ $cos~2A = \frac{cot~A-tan~A}{csc~A~sec~A}$
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