Chapter 5.4, Problem 87E

(a)

To determine

To represent:Thetrigonometric expression $\sin \theta +\cos \theta$ into the form of $\sqrt{a^2+b^2}\sin \left(B\theta +C\right)$ and verify using graph.

Answer to Problem 87E

The equation $\sin \theta +\cos \theta$ can be represented as $\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)$ .

Explanation of Solution

Given information:

The given equationis $\sin \theta +\cos \theta$ .

Formula Used:

  $C={\tan }^{-1}\left(\frac{b}{a}\right)$

Proof:

Compare the expression $\sin \theta +\cos \theta$ with $\sqrt{a^2+b^2}\sin \left(B\theta +C\right)$ .

  $a=1,b=1,B=1$

Calculate the value of $\sqrt{a^2+b^2}$ .

  $\begin{array}{c}\sqrt{a^2+b^2}=\sqrt{1^2+1^2}\\ =\sqrt{2}\end{array}$

Calculate the value of $C$ .

  $\begin{array}{c}C={\tan }^{-1}\left(\frac{b}{a}\right)\\ ={\tan }^{-1}\left(\frac{1}{1}\right)\\ =\frac{\pi }{4}\end{array}$

Consider the left hand side of the equation.

  $\begin{array}{c}\sin \theta +\cos \theta =\sqrt{a^2+b^2}\sin \left(B\theta +C\right)\\ =\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)\end{array}$

Draw the graph for the equation $\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)$ and $\sin \theta +\cos \theta$ using graphical utility.

  

Figure (1)

From the above, it is clear that both the graphs coincides. Therefore, the equation $\sin \theta +\cos \theta$ can be transformed into $\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)$ .

(b)

To determine

To represent:The trigonometric expression $\sin \theta +\cos \theta$ into the form of $\sqrt{a^2+b^2}\cos \left(B\theta -C\right)$ and verify using graph.

Answer to Problem 87E

The equation $\sin \theta +\cos \theta$ can be represented as $\sqrt{2}\cos \left(\theta -\frac{\pi }{4}\right)$ .

Explanation of Solution

Given information:

The given equationis $\sin \theta +\cos \theta$ .

Formula Used:

  $\begin{array}{l}{a}^2+b^2=r^2{\cos }^{2}C+r^2{\sin }^{2}C=r^2\left({\sin }^{2}C+{\cos }^{2}C\right)=r^2\\ \sqrt{a^2+b^2}=r^2\end{array}$

Proof:

Compare the expression $\sin \theta +\cos \theta$ with $\sqrt{a^2+b^2}\cos \left(B\theta -C\right)$ .

  $a=1,b=1,B=1$

Calculate the value of $\sqrt{a^2+b^2}$ .

  $\begin{array}{c}\sqrt{a^2+b^2}=\sqrt{1^2+1^2}\\ =\sqrt{2}\end{array}$

Calculate the value of $C$ .

  $\begin{array}{c}C={\tan }^{-1}\left(\frac{a}{b}\right)\\ ={\tan }^{-1}\left(\frac{1}{1}\right)\\ =\frac{\pi }{4}\end{array}$

Consider the left hand side of the equation.

  $\begin{array}{c}\sin \theta +\cos \theta =\sqrt{a^2+b^2}\cos \left(B\theta -C\right)\\ =\sqrt{2}\cos \left(\theta -\frac{\pi }{4}\right)\end{array}$

Draw the graph for the equation $\sqrt{2}\cos \left(\theta -\frac{\pi }{4}\right)$ and $\sin \theta +\cos \theta$ using graphical utility.

  

From the above, it is clear that both the graphs coincides. Therefore, the equation $\sin \theta +\cos \theta$ can be transformed into $\sqrt{2}\cos \left(\theta -\frac{\pi }{4}\right)$ .