Courses
Courses for Kids
Free study material
Offline Centres
More

Find the value of $\csc \left( {\dfrac{{11\pi }}{6}} \right)$.

seo-qna
Last updated date: 29th Feb 2024
Total views: 341.7k
Views today: 5.41k
IVSAT 2024
Answer
VerifiedVerified
341.7k+ views
Hint:We know that$\csc x = \dfrac{1}{{\sin x}}$. So first we need to find $\sin \left( {\dfrac{{11\pi }}{6}}
\right)$and then find its reciprocal.
We also know$\;\cos 2\theta = 1 - 2{\sin ^2}\theta $, this is one of the basic trigonometric identities.
In order to solve this question we can use the above mentioned identity. For that we have to convert our question in such a way that it can be expressed in the form of the above given identity.

Complete step by step solution:
Given
$\csc \left( {\dfrac{{11\pi }}{6}} \right)...................................\left( i \right)$
Now we also know$\csc x = \dfrac{1}{{\sin x}}$. So to find the value of $\csc \left( {\dfrac{{11\pi }}{6}}
\right)$we need to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$and then find it’s reciprocal.
Now to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$we can use the identity$\;\cos 2\theta = 1 - 2{\sin
^2}\theta $.
Finding the value of$\sin \left( {\dfrac{{11\pi }}{6}} \right)$:
Now let’s assume $\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a......................\left( {ii} \right)$
So similarly we can write $\;\cos a = \cos \left( {\dfrac{{11\pi }}{6}} \right)$
$ \Rightarrow \cos 2a = \cos \left( {\dfrac{{22\pi }}{6}} \right)$
We have to find the value of $\cos \left( {\dfrac{{22\pi }}{6}} \right)$such that by using the identity we can then solve the question.
So finding the value of$\cos \left( {\dfrac{{22\pi }}{6}} \right)$:
We know that $\cos \left( {\dfrac{{22\pi }}{6}} \right)$can be written as
$
\cos \left( {\dfrac{{12\left( {2\pi } \right)}}{6} - \dfrac{{2\pi }}{6}} \right) = \cos \left( {2\left( {2\pi }
\right) - \dfrac{{2\pi }}{6}} \right).................(iii) \\
\\
$
So from (iii) we know that \[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right)\]would be in
the IV Quadrant where cosine is positive.
Such that:

\[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{{2\pi }}{6}}
\right)..................(iv)\]
Also we know \[\cos \left( {\dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) =
\dfrac{1}{2}....................(v)\]
Now by using the identity $\;\cos 2\theta = 1 - 2{\sin ^2}\theta $we get
$
\Rightarrow \cos 2a = 1 - 2{\sin ^2}a = \dfrac{1}{2} \\
\Rightarrow 2{\sin ^2}a = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\
\Rightarrow {\sin ^2}a = \dfrac{1}{4} \\
$
Now from (ii) \[\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a\]
$
\Rightarrow \sin {\left( {\dfrac{{11\pi }}{6}} \right)^2} = \dfrac{1}{4} \\
\Rightarrow \sin \left( {\dfrac{{11\pi }}{6}} \right) = - \dfrac{1}{2} \\
$
Now we know that$\csc x = \dfrac{1}{{\sin x}}$, such that:
$\csc \left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{11\pi }}{6}} \right)}} =
\dfrac{1}{{\left( { - \dfrac{1}{2}} \right)}} = - 2$
Therefore the value of$\csc \left( {\dfrac{{11\pi }}{6}} \right)\;{\text{is}}\; - 2$.

Note:
General things to be known for solving this question.
I Quadrant:$0\; - \;\dfrac{\pi }{2}$ All values are positive.
II Quadrant:$\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.
III Quadrant:$\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.
IV Quadrant:$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.
Some other equations needed for solving these types of problem are:
\[
\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right) \\
\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right)-{\sin ^2}\left( \theta \right) = 1-2{\text{
}}{\sin ^2}\left( \theta \right) = 2{\text{ }}{\cos ^2}\left( \theta \right)-1 \\
\]

Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly.
Trending doubts