Question

How do you evaluate $\sec {{110}^{\circ }}$?

Answer 1

The secant function is equal to the reciprocal of the cosine function. So we can write the given trigonometric expression as $\dfrac{1}{\cos {{110}^{\circ }}}$. Then, we have to use the trigonometric identity given by $\cos 3x=4{{\cos }^{3}}x-3\cos x$ and substitute $x={{110}^{\circ }}$ in the identity. On the LHS, we will get $\cos {{330}^{\circ }}$ which will be equal to $\dfrac{\sqrt{3}}{2}$ using the trigonometric identity $\cos \left( {{360}^{\circ }}-x \right)=\cos x$. On putting $\cos {{110}^{\circ }}=x$, we will get a cubic equation in terms of $x$ which can be solved using the graphing calculator to get the value of $\cos {{110}^{\circ }}$. Putting the value of $\cos {{110}^{\circ }}$ in the expression $\dfrac{1}{\cos {{110}^{\circ }}}$, we will get the final answer.

Complete step by step solution:
Let us consider the trigonometric expression given in the above question as
$\Rightarrow E=\sec {{110}^{\circ }}$
Now, we know that secant is the reciprocal of the cosine function, that is, $\sec x=\dfrac{1}{\cos x}$. Therefore, the above trigonometric expression can also be written as
$\Rightarrow E=\dfrac{1}{\cos {{110}^{\circ }}}........\left( i \right)$
So we have to evaluate the expression $\cos {{110}^{\circ }}$. For this we consider the trigonometric identity given by
$\Rightarrow \cos 3x=4{{\cos }^{3}}x-3\cos x$
Substituting $x={{110}^{\circ }}$ in the above identity, we get

& \Rightarrow \cos 3\left( {{110}^{\circ }} \right)=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }} \\\ & \Rightarrow \cos {{330}^{\circ }}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }} \\\ \end{aligned}$$ Writing $${{330}^{\circ }}={{360}^{\circ }}-{{30}^{\circ }}$$ in the LHS, we get $$\Rightarrow \cos \left( {{360}^{\circ }}-{{30}^{\circ }} \right)=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}$$ Now, we know that $\cos \left( {{360}^{\circ }}-x \right)=\cos x$. So the above expression becomes $$\Rightarrow \cos {{30}^{\circ }}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}$$ We know that $$\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$$. Putting this in the above equation, we get $$\Rightarrow \dfrac{\sqrt{3}}{2}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}$$ Now let us substitute $$\cos {{110}^{\circ }}=x$$ in the above equation to get $$\Rightarrow \dfrac{\sqrt{3}}{2}=4{{x}^{3}}-3x$$ Subtracting $$\dfrac{\sqrt{3}}{2}$$ from both the sides, we get $$\begin{aligned} & \Rightarrow \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow 0=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow 4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2}=0 \\\ \end{aligned}$$ Considering the graph of $y=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2}$, we have  From the above graph, we get the roots as $x=-0.64,-0.34,0.98$. Since $x=\cos {{110}^{\circ }}$, the values $x=-0.34$ and $x=0.98$ are rejected. So we have $$\begin{aligned} & \Rightarrow x=-0.64 \\\ & \Rightarrow \cos {{110}^{\circ }}=-0.64 \\\ \end{aligned}$$ Putting this in (i) we get $\begin{aligned} & \Rightarrow E=\dfrac{1}{-0.34} \\\ & \Rightarrow E=-2.94 \\\ \end{aligned}$ Hence, the value of $\sec {{110}^{\circ }}$ is equal to $-2.94$. **Note:** After solving the cubic equation obtained above, do not conclude it as your final answer. Since we had put $$\cos {{110}^{\circ }}=x$$, the solution of the cubic equation will give the value of $$\cos {{110}^{\circ }}$$. But we have to find the value of $$\sec {{110}^{\circ }}$$ which is equal to $$\dfrac{1}{\cos {{110}^{\circ }}}$$.