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 \[-1\le \sin x\le 1\And -1\le \cos x\le 1\]. 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:
\[\begin{align}
  & \Rightarrow \left| \sin x \right|\le 1 \\
 & \Rightarrow -1\le \sin x\le 1 \\
 & \Rightarrow -1\le \sin \left( x+\dfrac{\pi }{4} \right)\le 1 \\
 & \Rightarrow \sqrt{2}\times \left( -1 \right)\le \sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)\le \sqrt{2}\times 1 \\
 & \Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)\le \sqrt{2} \\
\end{align}\]
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 \[\sin x+\cos x\]
\[\begin{align}
  & \Rightarrow Domain:\left( x+\dfrac{\pi }{4} \right)\in \left( -\infty ,\infty \right)\to x\in \left( -\infty ,\infty \right) \\
 & \Rightarrow Range:\left[ -\sqrt{2},\sqrt{2} \right] \\
\end{align}\]

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.