Trigonometric Ratios - SS1 Mathematics Lesson Note

The basic trigonometric ratios are defined in term of the sides of a right angles triangle.

Side \(a\) or \(BC\) is the opposite (opposite \(\theta\)).

Side \(b\) or \(AC\) is the adjacent (adjacent/beside \(\theta\)).

Side \(c\) or \(AB\) is the hypotenuse.

1. The basic trigonometric ratios \(sine\ \theta,\ cosine\ \theta\) and \(tangent\ \theta\) are defined as follows:

\(sine\ \theta = \ \frac{opposite}{hypotensue} = \frac{BC}{AB} = \frac{a}{c}\) \(cosine\ \theta = \ \frac{adjacent}{hypotensue} = \frac{AC}{AB} = \frac{b}{c}\) \(tangent\ \theta = \ \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{a}{b}\)

2. The reciprocal of these basic trigonometric functions are similarly important:

\[cosecant\ \theta\ (cosec\ \theta) = \frac{1}{\sin\theta} = \ \frac{hypotensue}{opposite} = \frac{AB}{BC} = \frac{c}{a}\]

\[secant\ \theta\ (sec\ \theta) = \frac{1}{\cos\theta} = \ \frac{hypotensue}{adjacent} = \frac{AB}{AC} = \frac{c}{b}\]

\[cotangent\ \theta\ (cot\ \theta) = \frac{1}{\tan\theta} = \ \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{b}{a}\]

3. Notice something important when considering \(sine\), \(cosine\) and \(tangent\):

\(\frac{\sin\theta}{\cos\theta} = \frac{\frac{a}{c}}{\frac{b}{c}} = \frac{a}{c} \div \frac{b}{c} = \frac{a}{c} \times \frac{c}{b} = \frac{a}{b} = \tan\theta\) And \(\frac{\cos\theta}{\sin\theta} = \frac{\frac{b}{c}}{\frac{a}{c}} = \frac{b}{c} \div \frac{a}{c} = \frac{b}{c} \times \frac{c}{a} = \frac{b}{a} = \cot\theta\)

4. Finally, the sine of any angle is equal to the cosine of its complementary angle and vice versa.

\(\sin\theta = \cos{(90{^\circ} - \theta)}\) \(\cos\theta = \sin{(90{^\circ} - \theta)}\) \(\tan\theta = \cot{(90{^\circ} - \theta)}\)

 

Example:

In a right-angled triangle where \(\tan\theta = \frac{4}{3}\). Prove that \(\cos\theta\sec\theta = 1\)

Solution

If \(\tan\theta = \frac{4}{3}\)

By Pythagoras theorem, \(c^{2} = a^{2} + b^{2}\)

\[c^{2} = 4^{2} + 3^{2}\]

\[c^{2} = 16 + 9\]

\[c^{2} = 25\]

\[c = \sqrt{25} = 5\]

Hence, \(\sin\theta = \frac{4}{5}\), \(\cos\theta = \frac{3}{5}\). \(sec\ \theta = \frac{1}{\cos\theta} = \frac{5}{3}\)

Therefore, \(\cos\theta\sec\theta = \ \frac{3}{5} \times \frac{5}{3} = 1\)

 

Example: Find the value of \(\theta\) in the equations: (i) \(\sin\theta - \cos\theta = 0\) (ii) \(\sin\theta = \cos 25{^\circ}\)

Solution

(i) \(\sin\theta - \cos\theta = 0\ \rightarrow \ \sin\theta = \cos\theta\ \rightarrow \ \sin\theta = \sin(90{^\circ} - \theta)\)

Hence, \(\theta = 90{^\circ} - \theta\ \rightarrow \ \theta + \theta = 90{^\circ}\ \rightarrow 2\theta = 90{^\circ}\ \rightarrow \ \theta = \frac{90{^\circ}}{2} = 45{^\circ}\)

(ii) \(\sin\theta = \cos 25{^\circ}\ \rightarrow \cos{(90{^\circ} - \theta) =}\cos 25{^\circ}\)

\[90{^\circ} - \theta = 25{^\circ}\ \rightarrow \ 90{^\circ} - 25{^\circ} = \theta\ \rightarrow 65{^\circ} = \theta\ \rightarrow \ \theta = 65{^\circ}\]

There are certain trigonometric ratios of angles that are special because they occur often in the real world. These ratios are compiled in the table below:

\[\mathbf{\theta}\]	\[\mathbf{0{^\circ}}\]	\[\mathbf{30{^\circ}}\]	\[\mathbf{45{^\circ}}\]	\[\mathbf{60{^\circ}}\]	\[\mathbf{90{^\circ}}\]
\[\mathbf{\sin}\mathbf{\theta}\]	\[0\]	\[\frac{1}{2}\]	\[\frac{\sqrt{2}}{2}\]	\[\frac{\sqrt{3}}{2}\]	\[1\]
\[\mathbf{\cos}\mathbf{\theta}\]	\[1\]	\[\frac{\sqrt{3}}{2}\]	\[\frac{\sqrt{2}}{2}\]	\[\frac{1}{2}\]	\[0\]
\[\mathbf{\tan}\mathbf{\theta}\]	\[0\]	\[\frac{\sqrt{3}}{2}\]	\[1\]	\[\sqrt{3}\]	\[\infty\]

\[90{^\circ} < \theta < 180{^\circ}\]

\[\sin{(180 - \theta)} = \sin\theta\]

\[\cos{(180 - \theta)} = - \cos\theta\]

\[\tan{(180 - \theta)} = - \tan\theta\]

\[\sin{180{^\circ}} = 0\]

\[\cos{180{^\circ}} = - 1\]

\[\tan{180{^\circ}} = 0\]

\[180{^\circ} < \theta < 270{^\circ}\]

\[\sin{(180 + \theta)} = - \sin\theta\]

\[\cos{(180 + \theta)} = - \cos\theta\]

\[\tan{(180 + \theta)} = \tan\theta\]

\[\sin{270{^\circ}} = - 1\]

\[\cos{270{^\circ}} = 0\]

\(\tan{270{^\circ}} = \infty\) (undefined)

\[270{^\circ} < \theta < 360{^\circ}\]

\[\sin{(360 - \theta)} = - \sin\theta\]

\[\cos{(360 - \theta)} = \cos\theta\]

\[\tan{(360 - \theta)} = - \tan\theta\]

\[\sin{360{^\circ}} = 0\]

\[\cos{360{^\circ}} = 1\]

\[\tan{360{^\circ}} = 0\]

Example: Without use of table, write down the value of each of the following in surd form:

1. \(\sin{135{^\circ}}\)

\(\cos{250{^\circ}}\)

\(\tan{330{^\circ}}\)

Solution

1. \(\sin{135{^\circ}} = \sin{(180 - 135)} = \sin{45{^\circ}} = \frac{\sqrt{2}}{2}\)

\(\cos{240{^\circ}} = \cos{(180 + 60)} = - \cos{60{^\circ}} = - \frac{1}{2}\)

\(\tan{330{^\circ}} = \tan{(360 - 330)} = - \tan{30{^\circ}} = - \frac{\sqrt{3}}{2}\)