Question

How do you find the exact values of \[\cos \left( {\dfrac{{11\pi }}{{12}}} \right)\]?

Answer 1

Hint:In this question, we can find our answer by using the basic rules and formulas of trigonometry. We can also use the half angle formula of \[\cos \]. After that check in which quadrant does the answer lie and then solve according to that.

Complete step by step answer:
The question Is to find the value of \[\cos \left( {\dfrac{{11\pi }}{{12}}} \right)\]. We will first assign a variable to \[\dfrac{{11\pi }}{{12}}\]. So, let’s say that \[\dfrac{{11\pi }}{{12}}\] is \[x\]. Then:
\[\cos \left( {\dfrac{{11\pi }}{{12}}} \right) = \cos x\] where \[x = \dfrac{{11\pi }}{{12}}\]. The equation can also be written as:
\[ \Rightarrow \cos x = \cos \left( {\dfrac{{11\pi }}{{12}}} \right)\]
If we multiply the equation with \[2\], then we get:
\[ \Rightarrow \cos 2x = \cos \left( {\dfrac{{11\pi }}{{12}}} \right) \times 2\]
\[ \Rightarrow \cos 2x = \cos \left( {\dfrac{{22\pi }}{{12}}} \right)\]

If we cancel out the divisible terms from the numerator and the denominator which are on the right side of the equation, then we get:
\[ \Rightarrow \cos 2x = \cos \left( {\dfrac{{11\pi }}{6}} \right)\]
\[ \Rightarrow \cos 2x = \cos \left( {\dfrac{\pi }{6}} \right)\]
\[ \Rightarrow \cos 2x = \dfrac{{\sqrt 3 }}{2}\]
Now, we will apply the trigonometry half angle formula. The formula is:
\[\cos 2x = 2{\cos ^2}x - 1\]
When we put the value of \[\cos 2x\] in the equation, then we get:
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = 2{\cos ^2}x - 1\]
This can also be written as:
\[ \Rightarrow 2{\cos ^2}x - 1 = \dfrac{{\sqrt 3 }}{2}\]

Now, we will shift \[1\] to the other side of the equation, then we get:
\[ \Rightarrow 2{\cos ^2}x = \dfrac{{\sqrt 3 }}{2} + 1\]
\[ \Rightarrow 2{\cos ^2}x = \dfrac{{\sqrt 3 + 2}}{2}\]
Now, we will shift \[2\] from the left side to the right side of the equation. When it gets shifted to the right, it gets divided and looks like:
\[ \Rightarrow {\cos ^2}x = \dfrac{{\sqrt 3 + 2}}{{2 \times 2}}\]
\[ \Rightarrow {\cos ^2}x = \dfrac{{\sqrt 3 + 2}}{4}\]
Now, we will get rid of the square on the left side. This is done by square rooting the right side and it looks like:
\[ \Rightarrow \cos x = \sqrt {\dfrac{{\sqrt 3 + 2}}{4}} \]
\[ \Rightarrow \cos x = \pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]

Initially we had taken \[\cos x\]as \[\cos \left( {\dfrac{{11\pi }}{{12}}} \right)\]. Now, when we put the value of \[\cos x\] in the equation, then we get:
\[ \Rightarrow \cos \left( {\dfrac{{11\pi }}{{12}}} \right) = \pm \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]
We know that \[\dfrac{{11\pi }}{{12}}\] comes under the 2nd Quadrant. Therefore, the answer will be negative. So, the answer we get is:
\[ \Rightarrow \cos \left( {\dfrac{{11\pi }}{{12}}} \right) = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]
If we check the calculator, then we will see that:
\[Arc\left( {\dfrac{{11\pi }}{{12}}} \right) = {165^ \circ }\]
 \[ \Rightarrow \cos \left( {\dfrac{{11\pi }}{{12}}} \right) = \cos {165^ \circ }\]
\[ \Rightarrow \cos \left( {\dfrac{{11\pi }}{{12}}} \right) = - 0.97\]
\[ \therefore - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} = - 0.97\]

Therefore, the exact value of \[\cos \left( {\dfrac{{11\pi }}{{12}}} \right)\] is \[ - 0.97\].

Note:The values of sine, cosine and tan in the First Quadrant are positive. The values of sine are only positive in the Second Quadrant. The values of tan are only positive in the third Quadrant and the values of cosine are only positive in the Fourth Quadrant.