Question

What is $\cot \left( \dfrac{\theta }{2} \right)$ in terms of trigonometric functions of a unit $\theta$?

Answer 1

This type of question depends on the concept of trigonometry. We use the relation between $\cot \left( \theta \right)$ and $\tan \left( \theta \right)$ that is $\cot \left( \theta \right)=\dfrac{1}{\tan \theta }$. Also, here we can use the basic definition of $\tan \theta$. Also we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\sin 2\theta =2\sin \theta \cos \theta$.

Complete step by step solution:
Now we have to express $\cot \left( \dfrac{\theta }{2} \right)$ in terms of trigonometric functions of a unit $\theta$.
For this let us consider,
$\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Let us multiply numerator as well denominator by $\sin \theta$
$\Rightarrow \tan \theta =\dfrac{{{\sin }^{2}}\theta }{\cos \theta \sin \theta }$
As we know that, ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$ we can write,
$\Rightarrow \tan \theta =\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta \sin \theta }$
By multiplying numerator and denominator by 2 we get,
$\Rightarrow \tan \theta =\dfrac{2-2{{\cos }^{2}}\theta }{2\cos \theta \sin \theta }$
We know that, $\sin 2\theta =2\sin \theta \cos \theta$
$\Rightarrow \tan \theta =\dfrac{1+1-2{{\cos }^{2}}\theta }{\sin 2\theta }$
Now we rearrange the numerator

& \Rightarrow \tan \theta =\dfrac{1-\left( -1 \right)-2{{\cos }^{2}}\theta }{\sin 2\theta } \\\ & \Rightarrow \tan \theta =\dfrac{1-\left( 2{{\cos }^{2}}\theta -1 \right)}{\sin 2\theta } \\\ \end{aligned}$$ As we know that, $$\cos 2\theta =2{{\cos }^{2}}\theta -1$$ we can write, $$\Rightarrow \tan \theta =\dfrac{1-\cos 2\theta }{\sin 2\theta }$$ By replacing $$\theta $$ with $$\left( \dfrac{\theta }{2} \right)$$ we get, $$\Rightarrow \tan \left( \dfrac{\theta }{2} \right)=\dfrac{1-\cos \theta }{\sin \theta }$$ Taking reciprocal of both the sides, we get, $$\Rightarrow \dfrac{1}{\tan \left( \dfrac{\theta }{2} \right)}=\dfrac{1}{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}$$ By the relation between trigonometric functions we have $$\cot \left( \theta \right)=\dfrac{1}{\tan \theta }$$ and hence, $$\Rightarrow \cot \left( \dfrac{\theta }{2} \right)=\dfrac{\sin \theta }{1-\cos \theta }$$ **Thus $$\dfrac{\sin \theta }{1-\cos \theta }$$is the representation of $$\cot \left( \dfrac{\theta }{2} \right)$$ in terms of trigonometric functions of a unit $$\theta $$.** **Note:** In this type of question students may make mistakes in understanding the question. Instead of writing in terms of the trigonometric function of a unit $$\theta $$ students may try to obtain an expression for $$\theta $$. Also students have to take care in the use of trigonometric formulae also. One of the students may directly obtain an expression for $$\cot \left( \dfrac{\theta }{2} \right)$$ by using the basic definition and the same formulae which also gives the same result. Also students have to take care during the conversion from a unit $$\theta $$ to half of $$\theta $$ that is $$\left( \dfrac{\theta }{2} \right)$$.