functions
Find the range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$.
By simply looking at the problem and simplifying trigonometrically,it looks as if range is zero but not.I think there is some trick to solve this.I am perplexed.I tried to find the domain first but not successful.Please help….
Given $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$
First we will calculate domain of function $f(x)$
function $\sin^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$ Similarly function $\cos^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$
and function $\tan^{-1}(x)$ is defined in $\displaystyle x\in \left(-\infty,+\infty\right)$
So $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$
So $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$ is defined in $$\displaystyle x\in \left[-1,1\right]$$
So $\displaystyle f(x) = \frac{\pi}{2}+\tan^{-1}(x)$
Now $\displaystyle f^{'}(x) = \frac{1}{1+x^2}>0\;\forall x\in [-1,1]$
So $f(x)$ is Strictly Increasing function.
So $\displaystyle f(-1) = \frac{\pi}{2}+\tan^{-1}(-1) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$
and $\displaystyle f(+1) = \frac{\pi}{2}+\tan^{-1}(1) = \frac{\pi}{2}+\frac{\pi}{4} = \frac{3\pi}{4}$
So $\displaystyle f(x)\in \left[\frac{\pi}{4}\;,\frac{3\pi}{4}\right]$
Geometric approach
Let's first look at the graphs of $\color{blue} \tan \color{blue}x$ and $\color{red} \cot \color{red} x$. (Note: the two graphs face each other in opposite directions)
Both tangent and cotangent have origin symmetry, and this means they are odd.
Algebraic approach
Foreknowledge: $\sin x$ is odd ($\sin(-x)=-\sin(x)$), $\cos x$ is even ($\cos(-x)=\cos(x)$).
$\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$
$\cot(-x)=\frac{1}{\tan(-x)}=\frac{\cos(-x)}{\sin(-x)}=\frac{\cos(x)}{-\sin(x)}=-\cot(x)$
Best Answer
Using $\displaystyle |\sin x|=\left\{\begin{matrix} \displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\ \displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\ \displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\ \displaystyle -\sin x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi \end{matrix}\right.$ and $\displaystyle |\cos x|=\left\{\begin{matrix} \displaystyle \cos x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\ \displaystyle -\cos x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\ \displaystyle -\cos x \;,& \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\ \displaystyle \cos x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi \end{matrix}\right.$
So $$\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}} = \sin x\cdot \left|\cos x\right|-\cos x\cdot \left|\sin x\right|$$
Here Function $f(x)$ is Periodic With Time Period $= 2\pi.$
So we will Calculate for Only one Time Period.
In $\bullet \displaystyle \; 0 \leq x\leq \frac{\pi}{2}\;,$ We get $f(x) = \sin x\cdot \cos x-\cos x\cdot \sin x = 0$
In $\bullet \displaystyle \; \frac{\pi}{2} \leq x\leq \pi\;,$ We get $f(x)=-\sin 2x .$ So $0 \leq f(x)\leq 1$
In $\bullet \displaystyle \; \pi \leq x\leq \frac{3\pi}{2}\;,$ We get $f(x)=0 .$
In $\bullet \displaystyle \; \frac{3\pi}{2} \leq x\leq 2\pi\;,$ We get $f(x)=\sin 2x .$ So $-1 \leq f(x)\leq 0$
So Here We get $\displaystyle -1 \leq f(x)\leq 1$