Proof for Symmetries in Trigo Identities

Let us see the properties of these trigo identities.
This section aims to help us find a shortcut to solving questions like: Use Compound angle formula to prove that $\cos \frac{19 π}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$ etc...

let us define $n$ as

n = k \times a + R

where $R$ is the remainder, and cannot be divided by $a$ . And $k, a, R, n \in Z$
For example

19 = 1 \times 12 + 7

Reducing $θ$ value

Expanding using the double angle formulae

\begin{aligned} \sin (\frac{n}{a} π) & = \sin (k π + \frac{R}{a} π) \\ = \cos (k π) \sin (\frac{R}{a} π) + \sin (k π) \cos (\frac{R}{a} π) \\ = \cos (k π) \sin (\frac{R}{a} π) \\ = (- 1)^{k} \sin (\frac{R}{a} π) \end{aligned}

Similarly

\begin{aligned} \cos (\frac{n π}{a}) & = \cos (k π) \cos (\frac{R}{a} π) \\ = (- 1)^{k} \cos (\frac{R}{a} π) \end{aligned}

Further reduction of $θ$

Since

\sin (π - x) = \sin x

\cos (π - x) = - \cos x

Thus

\sin (\frac{n}{a} π) = (- 1)^{k} \sin (\frac{a - R}{a} π)

\cos (\frac{n}{a} π) = (- 1)^{k + 1} \cos (\frac{a - R}{a} π)

Example

$\cos (\frac{19}{12} π) = \cos (π + \frac{7}{12} π) = \cos (π) \cos (\frac{7}{12} π) = - 1 \cdot \cos (\frac{7}{12} π)$

Then we can reduce $θ$ further

$- \cos \frac{7}{12} π = \cos \frac{12 - 7}{12} π = \cos \frac{5}{12} π$

Having reduced $θ$ . We can thus find the value of $\cos \frac{5}{12} π$ easily.

\begin{aligned} \cos \frac{5}{12} π & = \cos (\frac{π}{4} + \frac{π}{6}) \\ = \cos \frac{π}{4} \cos \frac{π}{6} - \sin \frac{π}{4} \sin \frac{π}{6} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{aligned}

Reducing θ value

Further reduction of θ

Reducing $θ$ value

Further reduction of $θ$