Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you evaluate \[\cos \left( \dfrac{13\pi }{12} \right)\] ?

seo-qna
Last updated date: 19th Jul 2024
Total views: 375.6k
Views today: 8.75k
Answer
VerifiedVerified
375.6k+ views
Hint: In this question we can find the value of the cosine of the given angle by following a few steps. Generally, we know that if we consider a right-angled triangle then the cosine of an angle is equal to the ratio of base to hypotenuse. In this first we will split \[\cos \left( \dfrac{13\pi }{12} \right)\] as sum of angles and then use identities to get \[-\cos \left( \dfrac{\pi }{12} \right)\] . Then, we will make use of trigonometric identity \[\cos \left( 2x \right)=2{{\cos }^{2}}x-1\] and obtain the value of cosx, i.e. \[\cos \left( \dfrac{\pi }{12} \right)\] from it. Thus, we will get the required value.

Complete step by step solution:
In the given question we have to evaluate \[\cos \left( \dfrac{13\pi }{12} \right)\] .
We can write \[\cos \left( \dfrac{13\pi }{12} \right)=\cos \left( \dfrac{\pi }{12}+\pi \right)\] as sum of angles as the cosine function repeats its value after one cycle. It is a periodic function.
We also know that
\[\begin{align}
  & \because \cos \left( \pi +\theta \right)=-\cos \theta \\
 & \therefore \cos \left( \dfrac{\pi }{12}+\pi \right)=-\cos \left( \dfrac{\pi }{12} \right) \\
\end{align}\]
Now, we have to evaluate the term \[\cos \left( \dfrac{\pi }{12} \right)\] . For that, we will make use of an identity given by \[\cos \left( 2x \right)=2{{\cos }^{2}}x-1\].
We have cosx as \[\cos \left( \dfrac{\pi }{12} \right)\] , so we will have cos2x as \[\cos \left( \dfrac{2\pi }{12} \right)\Rightarrow \cos \left( \dfrac{\pi }{6} \right)\Rightarrow \dfrac{\sqrt{3}}{2}\]. Now, substituting them, we get
\[\dfrac{\sqrt{3}}{2}=2{{\cos }^{2}}\left( \dfrac{\pi }{12} \right)-1\]
Shifting cos term to the LHS for ease of calculation, we will get
\[\begin{align}
  & 2{{\cos }^{2}}\left( \dfrac{\pi }{12} \right)-1=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow 2{{\cos }^{2}}\left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{3}}{2}+1 \\
\end{align}\]
Taking LCM of terms on the RHS, we get
\[\begin{align}
  & \Rightarrow 2{{\cos }^{2}}\left( \dfrac{\pi }{12} \right)=\dfrac{2+\sqrt{3}}{2} \\
 & \Rightarrow {{\cos }^{2}}\left( \dfrac{\pi }{12} \right)=\dfrac{2+\sqrt{3}}{4} \\
\end{align}\]
Now taking square root on both sides and neglecting the negative value because of the square root as \[\cos \left( \dfrac{\pi }{12} \right)\] lies in first quadrant which gives positive value, So we get
\[\begin{align}
  & \Rightarrow \cos \left( \dfrac{\pi }{12} \right)=\sqrt{\dfrac{2+\sqrt{3}}{4}} \\
 & \Rightarrow \cos \left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{2+\sqrt{3}}}{2} \\
\end{align}\]
Now, we had a relation as \[\cos \left( \dfrac{13\pi }{12} \right)=-\cos \left( \dfrac{\pi }{12} \right)\Rightarrow -\dfrac{\sqrt{2+\sqrt{3}}}{2}\].
Hence, we can conclude that \[\cos \left( \dfrac{13\pi }{12} \right)\] is \[-\dfrac{\sqrt{2+\sqrt{3}}}{2}\].

Note: While solving the above question we require trigonometric identities and formulas to evaluate so remember the basic trigonometric formulas for solving these types of questions. Also, keep in mind the domain and range of the cosine function while finding the value of \[\cos \left( \dfrac{13\pi }{12} \right)\] . Perform the calculations carefully. Keep in mind the procedure of solving this type of question.