Question

How do you derive exact algebraic formulae for $ \sin \left( \dfrac{\pi }{10} \right) $ and $ \cos \left( \dfrac{\pi }{10} \right) $ ?

Answer 1

Hint: We will write the angle as a sum of two angles. Then we will rearrange the equation and apply the sine function on both sides of that equation. We will use the multiple angle formulae for $ \cos 2x $ and $ \sin 3x $ . Then we will divide the polynomial equation to check the factors of the polynomial. Then we will use the quadratic formula to find the value of the asked sine function. Using this value, we will compute the value for the cosine function.

Complete step by step answer:
Let $ x=\dfrac{\pi }{10} $ . We can split the denominator as a product of 5 and 2. Then we have $ 5x=\dfrac{\pi }{2} $ . We will split the left hand side as a sum of two terms in the following manner,
 $ \begin{align}
  & 3x+2x=\dfrac{\pi }{2} \\
 & \therefore 3x=\dfrac{\pi }{2}-2x \\
\end{align} $
Now, we will take the sine function of the angles on both sides of the above equation.
 $ \sin 3x=\sin \left( \dfrac{\pi }{2}-2x \right) $
We know that $ \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ . Therefore, we have
 $ \sin 3x=\cos 2x $
Now, we have the following multiple angle formulae,
 $ \sin 3x=3\sin x-4{{\sin }^{3}}x $ ;
 $ \cos 2x=1-2{{\sin }^{2}}x $ .
Substituting these expressions in the above equation, we get
 $ 3\sin x-4{{\sin }^{3}}x=1-2{{\sin }^{2}}x $
Rearranging the above equation, we get
 $ 4{{\sin }^{3}}x-2{{\sin }^{2}}x-3\sin x+1=0 $
Now, we will divide the above polynomial by $ \sin x-1 $ in the following manner,
\[\begin{matrix}
   {} & 4{{\sin }^{2}}x & + & 2\sin x & - & 1 & {} & {} \\
   \sin x-1 & 4{{\sin }^{3}}x & - & 2{{\sin }^{2}}x & - & 3\sin x & + & 1 \\
   - & 4{{\sin }^{3}}x & - & 4{{\sin }^{2}}x & {} & {} & {} & {} \\
   {} & {} & {} & 2{{\sin }^{2}}x & - & 3\sin x & + & 1 \\
   - & {} & {} & 2{{\sin }^{2}}x & - & 2\sin x & {} & {} \\
   {} & {} & {} & {} & {} & -\sin x & + & 1 \\
   - & {} & {} & {} & {} & -\sin x & + & 1 \\
   {} & {} & {} & {} & {} & {} & {} & 0 \\
\end{matrix}\]
Therefore, we have the two factors as,
 $ \left( \sin x-1 \right)\left( 4{{\sin }^{2}}x+2\sin x-1 \right)=0 $
Dividing by $ \sin x-1 $ , we have $ 4{{\sin }^{2}}x+2\sin x-1=0 $ . We will use the quadratic formula to find the value of the sine function in the following manner,
 $ \begin{align}
  & \sin x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 4 \right)\left( -1 \right)}}{2\times 4} \\
 & \Rightarrow \sin x=\dfrac{-2\pm \sqrt{4+16}}{8} \\
 & \Rightarrow \sin x=\dfrac{-2\pm \sqrt{20}}{8} \\
 & \Rightarrow \sin x=\dfrac{-2\pm 2\sqrt{5}}{8} \\
 & \therefore \sin x=\dfrac{-1\pm \sqrt{5}}{4} \\
\end{align} $
Since, the angle $ x=\dfrac{\pi }{10} $ lies in the first quadrant, the sine function of this angle is positive. Therefore, we have $ \sin \left( \dfrac{\pi }{10} \right)=\dfrac{\sqrt{5}-1}{4} $ .
Now, we know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ and hence $ \cos x=\sqrt{1-{{\sin }^{2}}x} $ . Therefore, we have
 $ \cos \left( \dfrac{\pi }{10} \right)=\sqrt{1-{{\sin }^{2}}\left( \dfrac{\pi }{10} \right)} $
Substituting the obtained value of the sine function, we get
 $ \begin{align}
  & \cos \left( \dfrac{\pi }{10} \right)=\sqrt{1-{{\left( \dfrac{\sqrt{5}-1}{4} \right)}^{2}}} \\
 & \Rightarrow \cos \left( \dfrac{\pi }{10} \right)=\sqrt{1-\dfrac{6-2\sqrt{5}}{16}} \\
 & \Rightarrow \cos \left( \dfrac{\pi }{10} \right)=\sqrt{\dfrac{16-6-2\sqrt{5}}{16}} \\
 & \therefore \cos \left( \dfrac{\pi }{10} \right)=\dfrac{1}{4}\sqrt{10-2\sqrt{5}} \\
\end{align} $
So, the algebraic formulae are $ \sin \left( \dfrac{\pi }{10} \right)=\dfrac{\sqrt{5}-1}{4} $ and $ \cos \left( \dfrac{\pi }{10} \right)=\dfrac{1}{4}\sqrt{10-2\sqrt{5}} $ .

Note:
 It is important to understand trigonometric functions and their relations for such type of questions. We should be familiar with the formulae for trigonometric functions for multiple angles and the relations amongst themselves when adding or subtracting angles like $ \dfrac{\pi }{2},\pi $. These relations and formulae are essential in simplifying the given expressions and for calculations.