Trigonometric Ratios Trigonometry is used in many fields, such as construction, surveying, engineering, navigating and many other areas in research as experimental physics.

Trigonometry involves right-angled triangle extensively and there are ratios between the three sides that are useful in many ways. In order to refer to these ratios, the sides are named in relation to the angle being studied:

• the hypotenuse is the longest side, and is always opposite the right angle
• the side opposite to the studied angle marked ' $\theta$' (theta) is the opposite
• the adjacent side is next to the angle marked ' $\theta$'

The trigonometric ratios are: $\sin{\theta} =\cfrac { opposite }{ hypotenuse } \qquad \leftarrow Sine\\ \\ \cos{\theta} =\cfrac { adjacent }{ hypotenuse } \qquad \leftarrow Cosine\\ \\ \tan{\theta} =\cfrac { opposite }{ adjacent } \qquad \quad \leftarrow Tangent$

You may have seen these ratios in your scientific calculators. To remember these ratios easily, use the abbreviations: SOH CAH TOA.

SOH = Sine Opposite Hypotenuse

CAH Cosine Adjacent Hypotenuse

TOA Tangent Opposite Adjacent

As well as the basic ratios, there are invese trigonometric ratios: $\csc{\theta} =\cfrac { 1 }{ \sin{\theta} } =\cfrac { hypotenuse }{ opposite } \qquad \leftarrow Cosecant\\ \\ \sec{\theta} =\cfrac { 1 }{ \cos{\theta} } =\cfrac { hypotenuse }{ adjacent } \qquad \quad \leftarrow Secant\\ \\ \tan{\theta} =\cfrac { 1 }{ \tan{\theta} } =\cfrac { adjacent }{ opposite } \qquad \qquad \leftarrow Cotangent$

It is also useful to understand the complementary angles in a right-angled triangle: In $\triangle ABC, \angle C=\theta, \angle A=90^\circ-\theta$ $\sin{\theta} =\cos{(90^{ \circ }-\theta )}\\ \\ \cos{\theta} =\sin{(90^{ \circ }-\theta )}\\ \\ \sec{\theta} =\csc{(90^{ \circ }-\theta )}\\ \\ \csc{\theta} =\sec{(90^{ \circ }-\theta )}\\ \\ \tan{\theta} =\cot{(90^{ \circ }-\theta )}\\ \\ \cot{\theta} =\tan{(90^{ \circ }-\theta )}$

Examples:

1. Find the exact values for cos x, tan x and cosec x.  $\cos{x}\quad =\quad \cfrac { adjacent }{ hypotenuse } \quad =\quad \cfrac { 12 }{ 13 } \\ \\ \tan{x}\quad =\quad \cfrac { opposite }{ adjacent } \quad =\quad \cfrac { 5 }{ 12 } \\ \\ \csc{x}\quad =\quad \cfrac { hypotenuse }{ opposite } \quad =\quad \cfrac { 13 }{ 5 }$

2. If $\tan{\theta} =\cfrac { 3 }{ 4 }$  find $\sin{\theta} \quad and \quad \sec{\theta}$. If $\tan{\theta} =\cfrac { 3 }{ 4 }$ then $\triangle DEF$ above represents the triangle. The length of hypotenuse DF could be foudnm by Pythagoras' Theorem. ${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { 3 }^{ 2 }+{ 4 }^{ 2 }={ c }^{ 2 }\\ 25={ c }^{ 2 }\\ c=5$

Hence, $\sin{\theta}=\quad \cfrac { 3 }{ 5 } \\ \\ \sec{\theta} =\cfrac { 5 }{ 4 }$

3. Simplify $\cfrac { \cot { 25 } +\tan { 65 } }{ \cot { 25 } }$ $=\cfrac { \tan { 65 } +\tan { 65 } }{ \tan { 65 } } \\ \\ =\cfrac { 2\tan { 65 } }{ \tan { 65 } } \\ \\ =\quad 2$

Exact Ratios

For the following two triangles, a set of trigonometric ratios are useful for quick calculations.  $\sin { 0^{ \circ } } =0\\ \\ \cos { 0^{ \circ } } =1\\ \\ \tan { 0^{ \circ } } =0$ $\sin { 30^{ \circ } } =\cfrac { 1 }{ 2 } \\ \cos { 30^{ \circ } } =\cfrac { \sqrt { 3 } }{ 2 } \\ \tan { 30^{ \circ } } =\cfrac { 1 }{ \sqrt { 3 } }$ $\sin { 45^{ \circ } } =\cfrac { 1 }{ \sqrt { 2 } } \\ \cos { 45^{ \circ } } =\cfrac { 1 }{ \sqrt { 2 } } \\ \tan { 45^{ \circ } } =1$ $\sin { 60^{ \circ } } =\cfrac { \sqrt { 3 } }{ 2 } \\ \cos { 60^{ \circ } } =\cfrac { 1 }{ 2 } \\ \tan { 60^{ \circ } } =\sqrt { 3 }$ $\\ \sin { 90^{ \circ } } =1\\ \\ \cos { 90^{ \circ } } =0\\ \\ \tan { 90^{ \circ } } =\infty$

Think: Why does tan 90° result in $\infty$?

1. Find the exact value of $3\cos { \cfrac { \pi }{ 6 } } -\csc { \cfrac { \pi }{ 4 } }$. $3\cos { \cfrac { \pi }{ 6 } } -\csc { \cfrac { \pi }{ 4 } } \\ \\ =\quad 3\cos { \cfrac { \pi }{ 6 } } -\cfrac { 1 }{ \sin { \cfrac { \pi }{ 4 } } } \\ \\ =\quad 3\left( \cfrac { \sqrt { 3 } }{ 2 } \right) -\cfrac { \sqrt { 2 } }{ 1 } \\ \\ =\quad \cfrac { 3\sqrt { 3 } }{ 2 } -\cfrac { 2\sqrt { 2 } }{ 2 } \\ \\ =\quad \cfrac { 3\sqrt { 3 } -2\sqrt { 2 } }{ 2 }$

2. Find the exact value of $\tan { \left( -\cfrac { 4\pi }{ 3 } \right) }$.

(Refer to Unit Circle) $\pi =\cfrac { 3\pi }{ 3 } \\ \\ -\cfrac { 4\pi }{ 3 } =-(\pi +\cfrac { \pi }{ 3 } )$
The angle is in the 2nd quadrant, and tan $\theta$ is negative in the 2nd quadrant. $\tan { \left( -\cfrac { 4\pi }{ 3 } \right) } \\ \\ =-\tan { [-(\pi +\cfrac { \pi }{ 3 } )] } \left[ -\left( \pi +\cfrac { \pi }{ 3 } \right) \right] \\ \\ =-\tan { \cfrac { \pi }{ 3 } } \\ \\ =-\sqrt { 3 }$