Question

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

Answer 1

Hint: 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 $$-1\le \sin x\le 1\mathrm{\&}-1\le \cos x\le 1$$. Along with this in this question we use $$\sin (A+B)=\sin A\cos B+\cos A\sin B$$ and $$\sin \left({\displaystyle \frac{\pi }{4}}\right)=\cos \left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}$$.

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,
$$\Rightarrow \sin x+\cos x=\sqrt{2}\left({\displaystyle \frac{\sin x+\cos x}{\sqrt{2}}}\right)$$
By separating the denominator we can write,
$$\Rightarrow \sin x+\cos x=\sqrt{2}({\displaystyle \frac{1}{\sqrt{2}}}\sin x+{\displaystyle \frac{1}{\sqrt{2}}}\cos x)$$
We know that, $$\sin \left({\displaystyle \frac{\pi }{4}}\right)=\cos \left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}$$ hence,
$$\Rightarrow \sin x+\cos x=\sqrt{2}(\sin x\cos {\displaystyle \frac{\pi }{4}}+\cos x\sin {\displaystyle \frac{\pi }{4}})$$
Also, we know that, $$\sin (A+B)=\sin A\cos B+\cos A\sin B$$
$$\Rightarrow \sin x+\cos x=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$
Hence, we rewrite the given function $$\sin x+\cos x$$ as $$\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$.
As we know that the function $$\sin x$$ is defined for all real values of $$x$$. Thus the domain of $$\sin x$$ is:
$$\Rightarrow x\in (-\mathrm{\infty },\mathrm{\infty })$$
In similar manner we can say that the function $$\sin x+\cos x=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$ is also defined for all values of $$(x+{\displaystyle \frac{\pi }{4}})$$ and hence for all values of $$x$$. Thus we can say that the domain of the function $$\sin x+\cos x=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$ is:
$$\Rightarrow (x+{\displaystyle \frac{\pi }{4}})\in (-\mathrm{\infty },\mathrm{\infty })\to x\in (-\mathrm{\infty },\mathrm{\infty })$$
Also we know that the absolute value of $$\sin x$$ can never be greater than 1. So we have:
$$\begin{array}{rl}& \Rightarrow |\sin x|\le 1\\ & \Rightarrow -1\le \sin x\le 1\\ & \Rightarrow -1\le \sin (x+{\displaystyle \frac{\pi }{4}})\le 1\\ & \Rightarrow \sqrt{2}\times (-1)\le \sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})\le \sqrt{2}\times 1\\ & \Rightarrow -\sqrt{2}\le \sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})\le \sqrt{2}\end{array}$$
Thus, the value of the function $$\sin x+\cos x=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$ lies from $$-\sqrt{2}$$ to $$\sqrt{2}$$.
Hence, the range of $$\sin x+\cos x=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$ is:
$$\Rightarrow \sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})\in [-\sqrt{2},\sqrt{2}]$$
$$\Rightarrow \sin x+\cos x\in [-\sqrt{2},\sqrt{2}]$$
Hence, the domain and range of $$\sin x+\cos x$$
$$\begin{array}{rl}& \Rightarrow Domain:(x+{\displaystyle \frac{\pi }{4}})\in (-\mathrm{\infty },\mathrm{\infty })\to x\in (-\mathrm{\infty },\mathrm{\infty })\\ & \Rightarrow Range:[-\sqrt{2},\sqrt{2}]\end{array}$$

Note: In this type of question students may make a conversion of $$\sin x+\cos x$$ into $$\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$. 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.