Chapter 14.5, Problem 8CYU

To determine

To calculate:The of $\theta$ lying in between $0^{\circ }$ and ${360}^{\circ }$ satisfying thegiven equation.

Answer to Problem 8CYU

  $0^{\circ }and{135}^{\circ }$

Explanation of Solution

Given Information: An Equation $\sin \theta +\cos \theta =1$ is given.

Formula used: $\sin (A+B)=\sin A\cos B+\cos A\sin B.$

  $\begin{array}{l}GivenEquationcanbeWrittenas,\\ \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}}.\\ \therefore \sin \theta \cos {45}^{\circ }+\cos \theta \sin {45}^{\circ }=\sin {45}^{\circ }.\\ \therefore \sin (\theta +{45}^{\circ })=\sin {45}^{\circ }\\ \therefore \theta +{45}^{\circ }={45}^{\circ }or{180}^{\circ }-{45}^{\circ }\\ \therefore \theta =0^{\circ }or{135}^{\circ }\end{array}$