Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 32


There is one value of $\theta$ as solution to the equation: $$\{0^\circ\}$$

Work Step by Step

$$\sec\frac{\theta}{2}=\cos\frac{\theta}{2}$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $\frac{\theta}{2}$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $\frac{\theta}{2}$, the inequality would be $$0^\circ\le\frac{\theta}{2}\lt180^\circ$$ Thus, the corresponding interval for $\frac{\theta}{2}$ is $[0^\circ,180^\circ)$. 2) Now we examine the equation: $$\sec\frac{\theta}{2}=\cos\frac{\theta}{2}$$ Here we have both cosine and secant functions. It would be beneficial if we can change $\sec\frac{\theta}{2}$ into a cosine function, using the identity $\sec x=\frac{1}{\cos x}$ Thus, $$\frac{1}{\cos\frac{\theta}{2}}=\cos\frac{\theta}{2}$$ ($\cos\frac{\theta}{2}\ne0$) Multiply both sides with $\cos\frac{\theta}{2}$: $$\cos^2\frac{\theta}{2}=1$$ $$\cos\frac{\theta}{2}=\pm1$$ With $\cos\frac{\theta}{2}=1$, over interval $[0^\circ,180^\circ)$, there is 1 value whose sine equals $1$, which are $\{0^\circ\}$ With $\cos\frac{\theta}{2}=-1$, over interval $[0^\circ,180^\circ)$ (which does not include $180^\circ$), there is no value whose sine equals $-1$. Combining 2 cases, only 1 value has been found out, meaning $$\frac{\theta}{2}=\{0^\circ\}$$ It follows that $$\theta=\{0^\circ\}$$ This is the solution set of the equation.
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