Chapter 7, Problem 119RE

To determine

To solve:the given equation on the interval $0\le \theta <2\pi$ .

Answer to Problem 119RE

  $\theta \in \frac{\pi }{2},\pi$ .

Explanation of Solution

Given:

The given expression is $\sin \theta -\cos \theta =1$ .

Calculation:

As the given equationis $\sin \theta -\cos \theta =1$

  $\begin{array}{l}{\left(\sin \theta -\cos \theta \right)}^2=1\left\{\text{squaring both sides}\right\}\\ \left({\sin }^2\theta -2\sin \theta \cos \theta +{\cos }^2\theta \right)=1\\ 1-2\sin \theta \cos \theta =1\{{\cos }^2\theta +si{n}^2\theta =1\}\\ 1-\sin 2\theta =1\{\text{as }\sin 2\theta =2\sin \theta \cos \theta \}\\ \Rightarrow -\sin 2\theta =0\\ \therefore 2\theta =k\pi \\ \therefore \theta =\frac{k\pi }{2}\\ \therefore \theta =\frac{\pi }{2},\pi ,\frac{3\pi }{2}\end{array}$

Check the value of $\theta$ .

  $\begin{array}{l}\text{At }\theta =\frac{\pi }{2}\\ \Rightarrow \sin \theta -\cos \theta =1\\ \Rightarrow \sin \frac{\pi }{2}-\cos \frac{\pi }{2}=1\\ \Rightarrow 1-0=1\left(\text{true}\right)\end{array}$

  $\begin{array}{l}\text{At }\theta =\pi \\ \Rightarrow \sin \theta -\cos \theta =1\\ \Rightarrow \sin \pi -\cos \pi =1\\ \Rightarrow 0-\left(-1\right)=1\left(\text{true}\right)\end{array}$

  $\begin{array}{l}\text{At }\theta =\frac{3\pi }{2}\\ \Rightarrow \sin \theta -\cos \theta =1\\ \Rightarrow \sin \frac{3\pi }{2}-\cos \frac{3\pi }{2}=1\\ \Rightarrow -1-0=1\left(\text{false}\right)\end{array}$

Hence, the solution of $\sin \theta -\cos \theta =1$ is $\theta \in \frac{\pi }{2},\pi$ .