Question

The value of $\sin \left(\theta \right)+\cos \left(\theta \right)$ will be greatest when

• A. $\theta =30°$
• B. $\theta =45°$
• C. $\theta =60°$
• D. $\theta =90°$

Answer 1

The correct option is B

$\theta =45°$

Explanation for the correct option:

Step 1. Find the critical point

The given trigonometric expression: $\sin \left(\theta \right)+\cos \left(\theta \right)$.

Let us assume that, $f\left(\theta \right)=\sin \left(\theta \right)+\cos \left(\theta \right)$.

Differentiate both sides of the equation with respect to $\theta$.

$$\begin{gathered} \frac{d}{d\theta }\left[f\left(\theta \right)\right]=\frac{d}{d\theta }\left[\sin \left(\theta \right)+\cos \left(\theta \right)\right] \\ \Rightarrow f\text{'}\left(\theta \right)=\cos \left(\theta \right)-\sin \left(\theta \right) \end{gathered}$$

For critical point $f\text{'}\left(\theta \right)=0$

$$\begin{gathered} \Rightarrow \cos \left(\theta \right)-\sin \left(\theta \right)=0 \\ \Rightarrow \sin \left(\theta \right)=\cos \left(\theta \right) \\ \Rightarrow \tan \left(\theta \right)=1 \\ \Rightarrow \theta =\frac{\pi }{4},\frac{5\pi }{4} \end{gathered}$$

Step 2. Find the point of maxima

For maxima $f\text{'}\text{'}\left(\theta \right)<0$

Differentiate both sides of the equation with respect to $\theta$.

$$\begin{gathered} \frac{d}{d\theta }\left[f\text{'}\left(\theta \right)\right]=\frac{d}{d\theta }\left[\cos \left(\theta \right)-\sin \left(\theta \right)\right] \\ \Rightarrow f\text{'}\text{'}\left(\theta \right)=-\sin \left(\theta \right)-\cos \left(\theta \right) \end{gathered}$$

Now, $f\text{'}\text{'}\left(\frac{\pi }{4}\right)=-\sin \left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{4}\right)$.

$$\begin{gathered} \Rightarrow f\text{'}\text{'}\left(\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ \Rightarrow f\text{'}\text{'}\left(\frac{\pi }{4}\right)=-\frac{2}{\sqrt{2}} \\ \Rightarrow f\text{'}\text{'}\left(\frac{\pi }{4}\right)=-\sqrt{2} \end{gathered}$$

Thus, $f\text{'}\text{'}\left(\frac{\pi }{4}\right)<0$.

Now, $f\text{'}\text{'}\left(\frac{5\pi }{4}\right)=-\sin \left(\frac{5\pi }{4}\right)-\cos \left(\frac{5\pi }{4}\right)$.

$$\begin{gathered} \Rightarrow f\text{'}\text{'}\left(\frac{5\pi }{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ \Rightarrow f\text{'}\text{'}\left(\frac{5\pi }{4}\right)=\frac{2}{\sqrt{2}} \\ \Rightarrow f\text{'}\text{'}\left(\frac{5\pi }{4}\right)=\sqrt{2} \end{gathered}$$

Thus, $f\text{'}\text{'}\left(\frac{5\pi }{4}\right)>0$

Thus, $f\left(\theta \right)$ has a maximum value at $\theta =\frac{\pi }{4}$.

Therefore, the value of $\sin \left(\theta \right)+\cos \left(\theta \right)$ will be greatest when $\theta =\frac{\pi }{4}=45°$.

Hence, the correct option is (B).