Question

What is the domain and range of $\sin x+\cos x$?

Answer 1

This type of question depends on the basic concept of trigonometry. We know that $\sin x$ and $\cos x$ are defined for all real values of $x$. Also the absolute value of both the functions can never be greater than 1 as $-1\le \sin x\le 1\And -1\le \cos x\le 1$. Along with this in this question we use $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\sin \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$.

Complete step by step solution:
Now, we have to find domain and range of $\sin x+\cos x$
For this consider,
$\Rightarrow \sin x+\cos x$
Now, multiply and divide by $\sqrt{2}$ we get,
$\Rightarrow \sin x+\cos x=\sqrt{2}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)$
By separating the denominator we can write,
$\Rightarrow \sin x+\cos x=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x \right)$
We know that, $\sin \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ hence,
$\Rightarrow \sin x+\cos x=\sqrt{2}\left( \sin x\cos \dfrac{\pi }{4}+\cos x\sin \dfrac{\pi }{4} \right)$
Also, we know that, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
$\Rightarrow \sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$
Hence, we rewrite the given function $\sin x+\cos x$ as $\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$.
As we know that the function $\sin x$ is defined for all real values of $x$. Thus the domain of $\sin x$ is:
$\Rightarrow x\in \left( -\infty ,\infty \right)$
In similar manner we can say that the function $\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$ is also defined for all values of $\left( x+\dfrac{\pi }{4} \right)$ and hence for all values of $x$. Thus we can say that the domain of the function $\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$ is:
$\Rightarrow \left( x+\dfrac{\pi }{4} \right)\in \left( -\infty ,\infty \right)\to x\in \left( -\infty ,\infty \right)$
Also we know that the absolute value of $\sin x$ can never be greater than 1. So we have:

& \Rightarrow \left| \sin x \right|\le 1 \\\ & \Rightarrow -1\le \sin x\le 1 \\\ & \Rightarrow -1\le \sin \left( x+\dfrac{\pi }{4} \right)\le 1 \\\ & \Rightarrow \sqrt{2}\times \left( -1 \right)\le \sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)\le \sqrt{2}\times 1 \\\ & \Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)\le \sqrt{2} \\\ \end{aligned}$$ Thus, the value of the function $$\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$$ lies from $$-\sqrt{2}$$ to $$\sqrt{2}$$. Hence, the range of $$\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$$ is: $$\Rightarrow \sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)\in \left[ -\sqrt{2},\sqrt{2} \right]$$ $$\Rightarrow \sin x+\cos x\in \left[ -\sqrt{2},\sqrt{2} \right]$$ Hence, the domain and range of $$\sin x+\cos x$$ $$\begin{aligned} & \Rightarrow Domain:\left( x+\dfrac{\pi }{4} \right)\in \left( -\infty ,\infty \right)\to x\in \left( -\infty ,\infty \right) \\\ & \Rightarrow Range:\left[ -\sqrt{2},\sqrt{2} \right] \\\ \end{aligned}$$ **Note:** In this type of question students may make a conversion of $$\sin x+\cos x$$ into $$\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$$. Students may consider the function as it is and try to find out the domain and range or one may directly divide by $$\sqrt{2}$$. Students have to take care during multiplication and division by $$\sqrt{2}$$ as well as in conversion also.