Question

How do you find $\sin \left( \dfrac{\pi }{12} \right)$ and $\cos \left( \dfrac{\pi }{12} \right)$?

Answer 1

Hint: We have been given two trigonometric functions sine of $\dfrac{\pi }{12}$ and cosine of $\dfrac{\pi }{12}$ whose values are to be calculated. Thus, we shall assume these trigonometric functions as two separate variables, x and y, in order to simplify the equations for further calculations while using the trigonometric properties. Then, we shall simultaneously solve the formed equations to find the variables, x and y, and correspondingly the trigonometric functions.

Complete step by step solution:
Given $\sin \left( \dfrac{\pi }{12} \right)$ and $\cos \left( \dfrac{\pi }{12} \right)$.
We know that the angle $\dfrac{\pi }{12}$ occurs in the first quadrant and the sine as well as the cosine functions both are positive in the first quadrant. Thus, we can simplify assuming these trigonometric functions as two separate positive variables.
Let $x=\sin \left( \dfrac{\pi }{12} \right)$ and let $y=\cos \left( \dfrac{\pi }{12} \right)$, where $x$ and $y$are two positive variables.
From the basic properties of trigonometric functions, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.$\Rightarrow {{\sin }^{2}}\dfrac{\pi }{12}+{{\cos }^{2}}\dfrac{\pi }{12}=1$
$\Rightarrow {{x}^{2}}+{{y}^{2}}=1$ ……………….. (1)
According to the formula for sine of a double angle, $\sin 2\theta =2\sin \theta .\cos \theta $.
We can also write $\sin \left( \dfrac{\pi }{6} \right)$ as $\sin \left( 2.\dfrac{\pi }{12} \right)$.
$\Rightarrow \sin \left( 2.\dfrac{\pi }{12} \right)=2\sin \left( \dfrac{\pi }{12} \right).\cos \left( \dfrac{\pi }{12} \right)$
$\Rightarrow \sin \left( \dfrac{\pi }{6} \right)=2\sin \left( \dfrac{\pi }{12} \right).\cos \left( \dfrac{\pi }{12} \right)$
Applying the value of $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ as well as substituting the values of the assumed variables x and y, we get
$\Rightarrow \dfrac{1}{2}=2xy$
$\Rightarrow 2xy=\dfrac{1}{2}$ ………………. (2)
Now, we shall add equations (1) and (2).
$\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=1+\dfrac{1}{2}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=\dfrac{3}{2}$
Here, we will apply the algebraic property of square of sum of two terms, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ where $a=x$ and $b=y$.
 $\Rightarrow {{\left( x+y \right)}^{2}}=\dfrac{3}{2}$
Square rooting both sides, we get
$\Rightarrow \sqrt{{{\left( x+y \right)}^{2}}}=\sqrt{\dfrac{3}{2}}$
$\Rightarrow x+y=\dfrac{\sqrt{6}}{2}$ …………………… (3)
Now, we shall subtract equation (2) from equation (1).
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=1-\dfrac{1}{2}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=\dfrac{1}{2}$
Here, we will apply the algebraic property of square of difference of two terms, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ where $a=y$ and $b=x$.
$\Rightarrow {{\left( y-x \right)}^{2}}=\dfrac{1}{2}$
We shall understand that $y>x$for angle $\dfrac{\pi }{12}$. This is because sine function is an increasing function from 0 to $\dfrac{\pi }{12}$ whereas cosine function is a decreasing function from 0 to $\dfrac{\pi }{12}$. These functions meet at angle $\dfrac{\pi }{4}$. Therefore, at $\dfrac{\pi }{12}$, cosine function is greater than sine function.
Square rooting both sides, we get
$\Rightarrow \sqrt{{{\left( y-x \right)}^{2}}}=\sqrt{\dfrac{1}{2}}$
$\Rightarrow y-x=\dfrac{\sqrt{2}}{2}$ …………………… (4)
Further, we shall add equations (3) and (4) to find the value of variable-y.
$\Rightarrow x+y+y-x=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}$
$\Rightarrow 2y=\dfrac{\sqrt{6}+\sqrt{2}}{2}$
Dividing both sides of equation by 2, we get
$\Rightarrow y=\dfrac{\sqrt{6}+\sqrt{2}}{4}$
Now, we shall subtract equations (3) and (4) to find the value of variable-y.
$\Rightarrow x+y-\left( y-x \right)=\dfrac{\sqrt{6}}{2}-\dfrac{\sqrt{2}}{2}$
$\Rightarrow x+y-y+x=\dfrac{\sqrt{6}-\sqrt{2}}{2}$
$\Rightarrow 2x=\dfrac{\sqrt{6}-\sqrt{2}}{2}$
Dividing both sides of equation by 2, we get
$\Rightarrow x=\dfrac{\sqrt{6}-\sqrt{2}}{4}$
Therefore, $\sin \left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos \left( \dfrac{\pi }{12} \right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}$.

Note:
Another method of solving this problem and finding the values of $\sin \left( \dfrac{\pi }{12} \right)$ and $\cos \left( \dfrac{\pi }{12} \right)$ is by using the half angle formulae of cosine function and sine function, $\sin \left( \dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos \theta }{2}}$ and $\cos \left( \dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1+\cos \theta }{2}}$. These formulae relate an angle to the half of that angle. We can use them because we know the values of $\sin \left( \dfrac{\pi }{6} \right)$ and $\cos \left( \dfrac{\pi }{6} \right)$.