Trigonometry

Trigonometry originates from the study of triangles, particularly, the right angled ones.

Pythagoras Theorem

Pythagoras states that the expression is always be true for any three sides of a right angle triangle, where c is the hypothenuse.

a^{2} + b^{2} = c^{2}

Trigonometry

\sin x = \frac{opp}{hyp}

\cos x = \frac{adj}{hyp}

\tan x = \frac{opp}{adj}

Trigonometric Identities

\sin^{2} (x) + \cos^{2} (x) = 1

1 + \cot^{2} (x) = \csc^{2} (x)

\tan^{2} (x) + 1 = \sec^{2} (x)

We obtained the second expression by dividing the first expression throughout by $\sin^{2} (x)$ and the third by dividing the first expression by $\cos^{2} (x)$

Sine rule

\frac{\sin a}{A} = \frac{\sin b}{B} = \frac{\sin c}{C}

Cos rule

c^{2} = a^{2} + b^{2} - 2 a b \cos C

Double angle formula

\sin (a \pm b) = \sin a \cos b \pm \sin b \cos a

\cos (a \pm b) = \cos a \cos b \mp \sin a \sin b

\tan (a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}

Harmonic Addition Theorem

Express $R \sin (θ + a)$ as $A \cos θ + B \sin θ$ .

\begin{aligned} R \sin (θ + a) & = R (\sin θ \cos a + \sin a \cos θ) \\ = R \cos a \sin θ + R \sin a \cos θ \end{aligned}

Thus

A = R \sin a

B = R \cos a

Reverse Factor Formulae

2 \sin A \cos B = \sin (A + B) + \sin (A - B)

2 \cos A \sin B = \sin (A + B) - \sin (A - B)

2 \cos A \cos B = \cos (A + B) + \cos (A - B)

- 2 \sin A \sin B = \cos (A + B) - \cos (A - B)

Additional Insights (Very useful)

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 $k, a, R, n \in Z$ .
Look at Proof for Symmetries in Trigo Identities. We thus get.

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

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

Joshua's approximation method. Estimating $c o s$ $s i n$ $t a n$ without using their respective functions on a calculator

Suppose we want to find the value of

\cos (\frac{19}{5})

Using the approximation that $π \approx \frac{22}{7}$ . And reducing it using the Proof for Symmetries in Trigo Identities formulas

\begin{aligned} \cos (\frac{19}{5}) & = \cos (\frac{19}{5} \frac{7}{22} π) \\ = \cos (\frac{133}{110} π) \\ = (- 1)^{1} \cos (\frac{23}{110} π) \end{aligned}

Next, having reduced $θ < \frac{π}{2}$ . using the Maclaurin Series expansion for $\cos$ up to the $x^{4}$ term.

\begin{aligned} - \cos (\frac{23}{110} π) & \approx - (1 - \frac{{(\frac{23}{110} π)}^{2}}{2} + \frac{{(\frac{23}{110} π)}^{4}}{4!}) \end{aligned}

We see that

- (1 - \frac{{(\frac{23}{110} π)}^{2}}{2} + \frac{{(\frac{23}{110} π)}^{4}}{4!}) = - 0.792012968067

and using a calculator

\cos (\frac{19}{5}) = - 0.790967711914

Comparing the two solution, my method is accurate up to 2dp with an error of $0.001$ :)

\cos (\frac{19}{5}) - - (1 - \frac{{(\frac{23}{110} π)}^{2}}{2} + \frac{{(\frac{23}{110} π)}^{4}}{4!}) = 0.001045256153

Pythagoras Theorem

Trigonometry

Trigonometric Identities

Sine rule

Cos rule

Double angle formula

Harmonic Addition Theorem

Reverse Factor Formulae

Additional Insights (Very useful)

Joshua's approximation method. Estimating cos sin tan without using their respective functions on a calculator

Joshua's approximation method. Estimating $c o s$ $s i n$ $t a n$ without using their respective functions on a calculator