Question

Find the value of $\cos \left( \dfrac{x}{2} \right)$, if $\tan x=\dfrac{5}{12}$ and x lies in the $III$ quadrant?
(1)$\dfrac{5}{\sqrt{13}}$
(2) $\dfrac{5}{\sqrt{26}}$
(3) $\dfrac{5}{\sqrt{13}}$
(4) $\sqrt{\dfrac{1}{26}}$

Answer 1

To solve these types of questions, we should know all the formulas of trigonometry and also we should know which angle will be positive in which plane and which angle will be negative. All the angles of sine, cosine, tangent, cotangent, secant, and cosecant will remain positive in the first plane.

Complete step by step answer:
In the Cartesian system, the Cartesian plane is divided into four planes by the intersection of x-axis and y-axis. In the first quadrant, the angle lies between $0{}^\circ <\theta <90{}^\circ$. In the second quadrant the angle lies between $90{}^\circ <\theta <180{}^\circ$. In the third quadrant, the angle lies between $180{}^\circ <\theta <270{}^\circ$. In the fourth quadrant, the angle lies between $270{}^\circ <\theta <360{}^\circ$. In the first quadrant, all the values of the angle of sine, cosine, tangent, cosecant, secant, cotangent remain positive. In the second quadrant, the angle of sine and cosecant remains positive and all other angles become negative. In the third quadrant, the angle of tangent and cotangent remain positive and all other angles become negative. In the fourth quadrant, the angle of cosine and secant remains constant and all other angles become negative.
In the above question, it is given that x lies in the third quadrant and the value of $\tan x=\dfrac{5}{12}$.
According to the formula of trigonometry, we know that
${{\sec }^{2}}x-{{\tan }^{2}}x=1$
$\Rightarrow {{\sec }^{2}}x=1+{{\tan }^{2}}x$…….eq(1)
On putting the value of $\tan x=\dfrac{5}{12}$ in eq(1), we get the following results
${{\sec }^{2}}x=1+{{\left( \dfrac{5}{12} \right)}^{2}}$

& \Rightarrow {{\sec }^{2}}x=1+\dfrac{25}{144} \\\ & \Rightarrow {{\sec }^{2}}x=\dfrac{169}{144} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \sec x=\sqrt{\dfrac{169}{144}} \\\ & \Rightarrow \sec x=\dfrac{13}{12} \\\ \end{aligned}$$ We know that in the third quadrant the value of secant and cosine is negative so the value becomes. $$\sec x=\dfrac{-13}{12}$$ And we also know that the reciprocal of secant gives us cosine so the value becomes. $$\cos x=\dfrac{-12}{13}$$ Now according to the formula of trigonometry $$\cos 2x=2{{\cos }^{2}}x-1$$ $$\Rightarrow \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$$ $$\Rightarrow \cos x+1=2{{\cos }^{2}}\dfrac{x}{2}$$……….eq(2) Now we will put the value of $$\cos x=\dfrac{-12}{13}$$ in eq(2) and the following results will be obtained $$\dfrac{-12}{13}+1=2{{\cos }^{2}}\dfrac{x}{2}$$ $$\Rightarrow \dfrac{-12+13}{13}=2{{\cos }^{2}}\dfrac{x}{2}$$ $$\Rightarrow \dfrac{1}{13}=2{{\cos }^{2}}\dfrac{x}{2}$$ $$\Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{1}{26}$$ $$\Rightarrow \cos \left( \dfrac{x}{2} \right)=\dfrac{1}{\sqrt{26}}$$ On calculating, and solving questions using different identities of trigonometry, we have obtained the value of $$\cos \left( \dfrac{x}{2} \right)=\dfrac{1}{\sqrt{26}}$$. **So, the correct answer is “Option 4”.** **Note:** sine, cosine, secant, cosecant, tangent, and cotangent are also known as six trigonometric functions and these functions can also be obtained by finding the ratios of the side of the triangle. For eg- in a right-angled triangle, the ratio of perpendicular and base is equal to the tangent of the angle.