Question

How do you use the half-angle formulas to determine the exact values of sine, cosine and tangent of the angle $\dfrac{\pi }{{12}}$ ?

Answer 1

In order to solve this question, we need to use the half angle formula to find the respective values of the trigonometric functions. We find the value of $x$ by equating $\dfrac{x}{2}$ with $\dfrac{\pi }{{12}}$. We place this value in the half angle formula and solve it further to get our required answers.

Formula used: ${\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos x}}{2}$,${\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 + \cos x}}{2}$,${\tan ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos x}}{{1 + \cos x}}$

Complete Step by Step Solution:
In this question, we are asked to find the values of the trigonometric functions sine, cosine and tangent for the angle $\dfrac{\pi }{{12}}$ using the half angle formula.
All values less than $\dfrac{\pi }{2}$ fall in the first quadrant of a trigonometry unit circle.
As all the trigonometric functions are positive in the first quadrant hence the trigonometric functions for the angle $\dfrac{\pi }{{12}}$ will also be positive as it falls in the first quadrant.
The half angle formula for sine is given as:
$\Rightarrow {\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos x}}{2}$
Angle $\dfrac{x}{2}$ here corresponds to $\dfrac{\pi }{{12}}$
Therefore, $\dfrac{x}{2} = \dfrac{\pi }{{12}}$
$\Rightarrow x = \dfrac{\pi }{{12}} \times 2 = \dfrac{\pi }{6}$
Placing this value of $x$ in the formula, we get:
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 - \cos \left( {\dfrac{\pi }{6}} \right)}}{2}$
On simplifying it further, we get:
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 - \cos {{30}^ \circ }}}{2}$
As we know that the value of $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Therefore, ${\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}$
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\dfrac{{2 - \sqrt 3 }}{2}}}{2} = \dfrac{{2 - \sqrt 3 }}{2} \times \dfrac{1}{2}$
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{4}$
Let us take square root of both sides:
$\Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right) = \sqrt {\dfrac{{2 - \sqrt 3 }}{4}}$
$\Rightarrow \sin \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{1}{2}\sqrt {2 - \sqrt 3 }$
Let us find the value for cosine:
The half angle formula for cosine is given as ${\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 + \cos x}}{2}$
Similarly, here also $x = \dfrac{\pi }{6}$
Therefore, ${\cos ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 + \cos \left( {\dfrac{\pi }{6}} \right)}}{2}$
On simplifying it further, we get:
$\Rightarrow {\cos ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 + \cos {{30}^ \circ }}}{2}$
As we know that the value of $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Therefore, ${\cos ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{2}$
$\Rightarrow {\cos ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\dfrac{{2 + \sqrt 3 }}{2}}}{2} = \dfrac{{2 + \sqrt 3 }}{2} \times \dfrac{1}{2}$
$\Rightarrow {\cos ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 + \sqrt 3 }}{4}$
Let us take square root of both sides:
$\Rightarrow \cos \left( {\dfrac{\pi }{{12}}} \right) = \sqrt {\dfrac{{2 + \sqrt 3 }}{4}}$
$\Rightarrow \cos \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{1}{2}\sqrt {2 + \sqrt 3 }$
Let us find the half-angle value for tangent:
The half – angle formula for tangent is given as: ${\tan ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos x}}{{1 + \cos x}}$
Now as we already know that $x = \dfrac{\pi }{6}$
Therefore, ${\tan ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos \left( {\dfrac{\pi }{6}} \right)}}{{1 + \cos \left( {\dfrac{\pi }{6}} \right)}}$
$\Rightarrow {\tan ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \cos {{30}^ \circ }}}{{1 + \cos {{30}^ \circ }}}$
As we know that the value of $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Therefore, ${\tan ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{{1 + \dfrac{{\sqrt 3 }}{2}}}$
On simplifying it further, we get:
$\Rightarrow {\tan ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\dfrac{{2 - \sqrt 3 }}{2}}}{{\dfrac{{2 + \sqrt 3 }}{2}}}$
$\Rightarrow {\tan ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{2} \times \dfrac{2}{{2 + \sqrt 3 }}$
$\Rightarrow {\tan ^2}\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}$
Taking square root on both sides of the equation, we get:
$\Rightarrow \tan \left( {\dfrac{\pi }{{12}}} \right) = \sqrt {\dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}}$

Thus $\sqrt {\dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}}$ is the required answer.

Note: Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot.
We may even consider a complete circle and divide it into four quadrants to help us understand our trigonometric identities as so:

When the whole turn around the circle is equal to $2\pi$, while a half circle is equal to $\pi$ . The different quadrants are divided into different angles. All trigonometric identities have positive signs in the first quadrant, while sine has positive values in the second quadrant. Tan has positive values in the third quadrant and cosine has positive values in the fourth quadrant.