Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 208: 32



Work Step by Step

$$A=\sin^3\alpha+\cos^3\alpha$$ Now it is crucial here not to forget that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ which means $$A=(\sin\alpha+\cos\alpha)(\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha)$$ $$A=(\sin\alpha+\cos\alpha)[(\sin^2\alpha+\cos^2\alpha)-\sin\alpha\cos\alpha]$$ - From Pythagorean Identity: $$\sin^2\alpha+\cos^2\alpha=1$$ So, $A$ would be $$A=(\sin\alpha+\cos\alpha)(1-\sin\alpha\cos\alpha)$$
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