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Introduction

Trigonometric Functions (on Angles)	Trigonometric Ratios (of Lengths)	Mnemonics
$\sin \theta $	\[\frac{{{\rm{Opposite}}}}{{{\rm{Hypotenuse}}}}\]	SOH
$\cos \theta $	\[\frac{{{\rm{Adjacent}}}}{{{\rm{Hypotenuse}}}}\]	CAH
$\tan \theta $	\[\frac{{{\rm{Opposite}}}}{{{\rm{Adjacent}}}}\]	TOA

Download the Notes (Mnemonics): http://db.tt/DaO9aP6Z

In the past, when calculators were not easily available, trigonometric functions/ratios were obtained from a "Trigonometric Table", as shown below.

Uses of Trigonometry

History of Trigonometry

Trigonometric Ratios of Acute Angles

Below shows ∠BAC.

You can shift the points B and C to change the size of ∠BAC.

Instructions:

Adjust the size of ∠BAC.
Check the boxes.
Observe the ratios of the lengths of the sides of the right-angled △ABC.
Observe the values of the trigonometric functions: $\sin \theta $, $\cos \theta $ and $\tan \theta $.
Repeat steps 1 ~ 4 for other sizes of ∠BAC.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/44996

Download the GeoGebra Worksheet: http://db.tt/qkrYqnYQ

Download the Worksheet: http://db.tt/ZWW78kyk

Trigonometric Ratios of Complementary Angles

Complementary Angles: Angles that add up to ${90^\circ }$.

Below shows right-angled △ABC.

You can shift the points B and C to change the size of △ABC.

Instructions:

Adjust the size of △ABC.
Check the boxes.
Observe the values of the trigonometric functions: $\sin \theta $, $\cos \theta $ and $\tan \theta $.
Observe the values of the trigonometric functions: $\sin \left( {90^\circ - \theta } \right)$, $\cos \left( {90^\circ - \theta } \right)$ and $\tan \left( {90^\circ - \theta } \right)$.
Repeat steps 1 ~ 4 for other sizes of △ABC.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/44997

Download the Worksheet: http://db.tt/y5PJ7T6P

Trigonometric Ratios of Obtuse Angles

Below shows ∠DAC.

You can shift the point C to change the size of ∠DAC.

Instructions:

Adjust the size of ∠DAC.
Check the boxes.
Observe the values of the trigonometric functions: $\sin \theta $, $\cos \theta $ and $\tan \theta $.
Observe the values of the trigonometric functions: $\sin \left( {180^\circ - \theta } \right)$, $\cos \left( {180^\circ - \theta } \right)$ and $\tan \left( {180^\circ - \theta } \right)$.
Repeat steps 1 ~ 4 for other sizes of ∠DAC.

Supplementary Angles: Angles that add up to ${180^\circ }$.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/44998

Download the GeoGebra Worksheet: http://db.tt/3guaUADv

Download the Worksheet: http://db.tt/6MFfkRFB

Area of Triangles

Below shows △ABC.

You can shift the points B and C to change the size of △ABC.

Instructions:

Adjust the size of △ABC.
Check the boxes.
Observe the area of △ABC obtained by both methods.
Compare both formulae.
Repeat steps 1 ~ 4 for other sizes of △ABC.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/44999

Download the Worksheet: http://db.tt/O5TJ2ZPM

Sine Rule

Below shows △ABC.

You can shift the points B and C to change the size of △ABC.

Instructions:

Adjust the size of △ABC.
Check the boxes.
Observe the values of the ratios.
Repeat steps 1 ~ 3 for other sizes of △ABC.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/45000

Download the Worksheet: http://db.tt/nY22xlli

Cosine Rule

Below shows △ABC.

You can shift the points B and C to change the size of △ABC.

Instructions:

Adjust the size of △ABC.
Check the boxes.
Observe the values of ${a^2}$ and ${b^2} + {c^2} - 2bc\cos A$.
Observe the values of $\cos A$ and $\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$.
Repeat steps 1 ~ 4 for other sizes of △ABC.

Download the GeoGebra file: http://www.geogebratube.org/material/download/format/file/id/45001

Download the Worksheet: http://db.tt/EgNEyXaF

Area of Triangles (Proof)

Proof:
\[Area\;of\;\Delta ABC = \frac{1}{2}ch\]
Since
\[\sin \theta = \frac{b}{h}\]
\[h = b\sin \theta \]
Thus,
\[Area\;of\;\Delta ABC = \frac{1}{2}cb\sin \theta \]
where $\theta $ is the angle between the sides $b$ and $c$.

Sine Rule (Proof)

Proof:
From the Trigonometric Formula for Areas of Triangles,
\[\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C\]
\[bc\sin A = ac\sin B = ab\sin C\]
\[\frac{{bc\sin A}}{{abc}} = \frac{{ac\sin B}}{{abc}} = \frac{{ab\sin C}}{{abc}}\]
\[\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{\sin C}}{c}\]
Thus,
\[\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\]
where $\angle A$ is the angle opposite the side $a$, $\angle B$ is the angle opposite the side $b$ and $\angle C$ is the angle opposite the side $c$.

Cosine Rule (Proof)

Proof:

Case 1: $\angle A = 90^\circ $

Since
$$\begin{align}
\cos A & = \cos 90^\circ \\
& = 0 \\
\end{align}$$
Then,
\[2bc\cos A = 0\]
By Pythagoras' Theorem,
$$\begin{align}
{a^2} & = {b^2} + {c^2} \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$

Case 2: $\angle A < 90^\circ $

Let $CP = h$ be the perpendicular height of $\Delta ABC$.
Let $AP = m$ and $BP = n$ such that $n = c - m$.
$$\begin{align}
\cos A & = \frac{m}{b} \\
m & = b\cos A \\
\end{align}$$
From $\Delta ACP$, by Pythagoras' Theorem,
\[{h^2} = {b^2} - {m^2}\]
From $\Delta BCP$, by Pythagoras' Theorem,
\[{h^2} = {a^2} - {\left( {c - m} \right)^2}\]
Therefore,
$$\begin{align}
{b^2} - {m^2} & = {a^2} - {\left( {c - m} \right)^2} \\
& = {a^2} - {c^2} + 2cm - {m^2} \\
{b^2} & = {a^2} - {c^2} + 2cm \\
{a^2} & = {b^2} + {c^2} - 2cm \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$

Case 3: $\angle A > 90^\circ $

Let $CP = h$ be the perpendicular height of $\Delta ABC$.
Let $AP = m$.
$$\begin{align}
\cos A & = - \cos \left( {180^\circ - A} \right) \\
& = - \frac{m}{b} \\
m & = - b\cos A \\
\end{align}$$
From $\Delta ACP$, by Pythagoras' Theorem,
\[{h^2} = {b^2} - {m^2}\]
From $\Delta BCP$, by Pythagoras' Theorem,
\[{h^2} = {a^2} - {\left( {c + m} \right)^2}\]
Therefore,
$$\begin{align}
{b^2} - {m^2} & = {a^2} - {\left( {c + m} \right)^2} \\
& = {a^2} - {c^2} - 2cm - {m^2} \\
{b^2} & = {a^2} - {c^2} - 2cm \\
{a^2} & = {b^2} + {c^2} + 2cm \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$