Question

Find the approximate value of $\tan {{46}^{\circ }}$ (take $\pi =\dfrac{22}{7}$). $$$$

Answer 1

We write the given tangent trigonometric argument as $\tan {{46}^{\circ }}=\tan \left( {{45}^{\circ }}+{{1}^{\circ }} \right)$.We convert the given measures in degree for the tangent function into radian using the conversion formula${{R}^{c}}=\dfrac{\pi }{180}\times {{D}^{\circ }}$. We use the working rule for approximating a function $f\left( x \right)$ for a small quantity change $\delta x$ using first order differential ${{f}^{'}}\left( x \right)$ as $f\left( x+\delta x \right)\simeq f\left( x \right)+{{f}^{'}}\left( x \right)\delta x$. We take $x={{45}^{\circ }},\delta x={{1}^{\circ }}$ in radians to proceed. $$$$

Complete step-by-step solution:
We know that from differential calculus that we can approximate the function $f\left( x+\delta x \right)$ as $f\left( x \right)+{{f}^{'}}\left( x \right)\delta x$ where ${{f}^{'}}\left( x \right)$ is the first derivative of the function $f\left( x \right)$ and $\delta x$ is a very small quantity. So we have;
$f\left( x+\delta x \right)\simeq f\left( x \right)+{{f}^{'}}\left( x \right)\delta x=f\left( x \right)+\delta x\dfrac{d}{dx}f\left( x \right)$
We know how to convert from given measure degree ${{D}^{\circ }}$to radian ${{R}^{c}}$ as follows
${{R}^{c}}=\dfrac{\pi }{180}\times {{D}^{\circ }}$
We know that the trigonometric function $\tan x$ takes its input in radians and from the set $\mathsf{\mathbb{R}}-\dfrac{2n+1}{2}\pi$ and returns from the set$\mathsf{\mathbb{R}}$.
We are asked in the question to find the approximate values of $\tan {{46}^{\circ }}$. Let us consider
$\tan {{46}^{\circ }}=\tan \left( {{45}^{\circ }}+{{1}^{\circ }} \right)$
We convert the measure to radian and have ;

& {{45}^{\circ }}={{\left( 45\times \dfrac{\pi }{180} \right)}^{c}}=\dfrac{{{\pi }^{c}}}{4} \\\ & {{1}^{\circ }}={{\left( 1\times \dfrac{\pi }{180} \right)}^{c}}={{\left( 1\times \dfrac{\dfrac{22}{7}}{180} \right)}^{c}}={{\left( \dfrac{22}{1260} \right)}^{c}}={{0.0176}^{c}} \\\ \end{aligned}$$ Now we can write $\tan {{46}^{\circ }}$ as $$\tan {{46}^{\circ }}=\tan \left( {{45}^{\circ }}+{{1}^{\circ }} \right)\simeq \tan \left( \dfrac{{{\pi }^{c}}}{4}+{{0.0176}^{c}} \right)$$ We see that $0.0176$ is a very small quantity compared to $\dfrac{\pi }{4}\simeq 0.7854$. So we can use differential approximation for small change $\delta x$ and have; $$\tan \left( x+\delta x \right)\simeq \tan x+\delta x\dfrac{d}{dx}\left( \tan x \right)$$ We know from standard differential of tangent function is the square of secant that is $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$. So we have; $$\Rightarrow \tan \left( x+\delta x \right)\simeq \tan x+\delta x{{\sec }^{2}}x$$ We take $x={{\left( \dfrac{\pi }{4} \right)}^{c}}$ and $\delta x\simeq {{0.01746}^{c}}$ in the above step to have to have; $$\Rightarrow \tan \left( \dfrac{\pi }{4}+0.01746 \right)\simeq \tan \left( \dfrac{\pi }{4} \right)+0.01746\times {{\sec }^{2}}\left( \dfrac{\pi }{4} \right)$$ We know that value of $\tan \left( \dfrac{\pi }{4} \right),\sec \left( \dfrac{\pi }{4} \right)$ trigonometric table as $\tan \left( \dfrac{\pi }{4} \right)=\tan {{45}^{\circ }}=1$ and $\sec \left( \dfrac{\pi }{4} \right)=\sec {{45}^{\circ }}=\sqrt{2}$ respectively. We put these values in the above step to have $$\begin{aligned} & \Rightarrow \tan \left( \dfrac{\pi }{4}+0.01746 \right)\simeq 1+0.01746\times {{\left( \sqrt{2} \right)}^{2}} \\\ & \Rightarrow \tan \left( \dfrac{\pi }{4}+0.01746 \right) \simeq 1+0.01746\times 2 \\\ & \Rightarrow \tan \left( {{46}^{\circ }} \right) \simeq 1+0.03492 \\\ & \Rightarrow \tan \left( {{46}^{\circ }} \right) \simeq 1.03492 \\\ \end{aligned}$$ **Note:** We note that the approximation formula comes from Taylor’s series which is given by $f\left( x+h \right)=f\left( x \right)+h{{f}^{'}}\left( x \right)+\dfrac{{{h}^{2}}}{2!}{{f}^{''}}\left( x \right)+\dfrac{{{h}^{3}}}{3!}{{f}^{'''}}\left( x \right)+...$ where $h$ is the very small change in the function $f\left( x \right)$.If we take $y=\tan x$ we can approximate small change in $y$ as $\delta y$ corresponding to small change $\delta x\simeq dx$ in $x$ as $\delta y\simeq dy=\dfrac{d}{dx}\left( y=\tan x \right)\times dx$. We should remember that all trigonometric functions take their arguments in radians not in degree.