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# Trigonometrical Ratios

Describes the derivation of the basic ratios and their relationship one with another

## Trigonometric Ratios For Any Angle.

A knowledge of the Pythagoras theorem and the properties of Similar triangles, is assumed.

## Definition Of The Basic Ratios.

Consider the two right angled triangles shown. As they are equiangular the following relationships exist between the lengths of their sides.

Diagram

$\frac{b}{y}&space;=&space;\frac{a}{x}$

Re-arranging:-

$\frac{b}{a}&space;=&space;\frac{y}{x}$

These two ratios are clearly independent of the size of the triangle and depend solely upon the size of the angles of a right angled triangle. By definition the value of equalities shown in equation (2) is called $\inline&space;&space;\b&space;Sin\;\theta$.

Using the same analysis two other ratios can be identified. These are $\inline&space;&space;\frac{c}{a}\;and;\frac{b}{c}\;and\;are\;called\;Cos\;\theta&space;\;and\;Tan\;\theta\;respectively$

In Conclusion, considering the following right angled triangle:-

$\mathbf{Sin\;\theta&space;&space;=&space;\frac{b}{a}}$

$\mathbf{Cos\;\theta&space;&space;=&space;\frac{c}{a}}$

$\mathbf{Tan\;\theta&space;&space;=&space;\frac{b}{c}}$

## Other Identities

For convenience the following identities exist:-

$\mathbf{Cosec\;\theta&space;&space;=&space;\frac{1}{Sin\;\theta&space;}}$

$\mathbf{Sec\;\theta&space;&space;=&space;\frac{1}{Cos\;\theta&space;}}$

$\mathbf{Cot\;\theta&space;&space;=&space;\frac{1}{Tan\;\theta&space;}}$

## The Relationships Between Sin; Cos And Tan.

By Inspection of equations (3:4:5)it can be seen that

$\mathbf{Tan\;\theta&space;&space;=&space;\frac{Sin\;\theta&space;}{Cos\;\theta&space;}}$

In addition if we put a = 1 . The following diagram can be drawn. The two perpendicular Coordinates are Ox and Oy. The radius OP is of unit length and the angle xOP is measured clockwise from xOP and the coordinates of P are defined as $\inline&space;&space;Cos\;\theta\;and\;Sin\;\theta$ whatever the position of P.

The signs are defined as in ordinary Algebraic graphs.

From the above Graph it can be seen that :-

$\mathbf{Sin^2\;\theta&space;&space;+&space;Cos^2\;\theta&space;&space;=&space;1}$

Dividing through by $\inline&space;&space;Cos^2\;\theta$

$\mathbf{1&space;&space;+&space;Tan^2\;\theta&space;&space;=&space;Sec^2\;\theta&space;}$

Or dividing by $\inline&space;&space;Sin^2\;\theta$ gives:-

$\mathbf{1&space;&space;+&space;Cot^2\;\theta&space;&space;=&space;Cosec^2\;\theta&space;}$

## Special Angles

From the definitions and from an inspection of the graph it is possible to rite down the Ratios for the following angles.

It can also be seen that Sin 45 = Cos 45 and that as the sum of their squares = 1

$\mathbf{Cos\;45^0&space;=&space;Sin\;45^0&space;=&space;\frac{1}{\sqrt{2}}}$

From the above diagrams it is possible to see that :-

$\mathbf{Cos\;60^0&space;=&space;\frac{1}{2}\;\;\;\;\;and\;\;\;\;\;Sin\;60^0&space;=&space;\frac{\sqrt{3}}{2}}$

Similarly

$\mathbf{Sin\;30^0&space;=&space;\frac{1}{2}\;\;\;\;\;and\;\;\;\;\;Cos\;30^0&space;=&space;\frac{\sqrt{3}}{2}}$

## Complementary Angles

If angles xOP and xOQ are complementary then Angles xOP and QOy are equal and so the projection of OP onto the x axis is equal to the projection of OQ onto the y-axis.

$\therefore\;\;\;\;\;Cos(90^0&space;-&space;\theta&space;)&space;=&space;sin\;\theta$

and similarly

$\therefore\;\;\;\;\;Sin(90^0&space;-&space;\theta&space;)&space;=&space;Cos\;\theta$

NOTE The prefix co- in the ratios stands for "complementary" and means that the ratio of any angle is equal to the co-ratio of the complementary angle. e.g.

$\mathbf{tan(90^0&space;-&space;\theta&space;)&space;=&space;cot\;\theta&space;}$

## Angles Larger Than 90 Degrees.

From the definitions and from the graph it can be seen that:-

• In the first quadrant, sin is +ve; cos is +ve; tan is +ve.
• In the second quadrant, sin is +ve; cos is -ve; tan is -ve.
• In the third quadrant, sin is -ve; cos is -ve; tan is +ve.
• In the forth quadrant, sin is -ve; cos is +ve; tan is -ve.

There are various methods of remembering the above table. One way is by using the CAST diagram.

Each letter stands for the positive ratios. e.g. In the first quadrant all ratios are positive whilst in the third quadrant only tan is positive.

Since the magnitude of the projections of the unit radius on the axies depend solely on the acute angle which the radius makes with the x-axis, any ratio of any angle is equal numerically to the same ratio of the acute angle which the radius makes with the x-axis. The sign must be found from the Cast circle.

Example

Find the value of sin 210 degrees.

As 210 is in the third quadrant the sin is negative and the acute angle which the radius makes with the x-axis is 30 degrees.

$\therefore\;\;\;\;\;sin\;210^0&space;=&space;sin(210&space;-&space;180)\;=&space;-&space;sin\;30\;=&space;-&space;\frac{1}{2}$

To simplify expressions such as $\inline&space;&space;tan(270^0&space;-&space;\theta)$ follow the same logic but for simplicity imagine that $\inline&space;&space;\theta$ to be acute. The formula will apply for all values of $\inline&space;&space;\theta$ but the argument is simplified by supposing that it is acute.

Examples

$sin(180^0&space;-&space;\theta&space;)\;=&space;+&space;sin\;\theta$

$cos(180^0&space;+&space;\theta&space;)\;=&space;-&space;cos\;\theta$

$tan(270^0&space;-&space;\theta&space;)\;=&space;+&space;tan\;(90^0\;-\theta)&space;=&space;cot\;\theta$

$sec(360^0&space;-&space;3\theta&space;)\;=&space;+&space;sec\;3\theta$

$cot(-&space;2\theta&space;)\;=&space;-&space;cot\;2\theta$

It should be noted that cosec; sec; and cot follow the same rules as sin;cos; and tan respectively since they are their reciprocals.

### Example 1

Prove the following identities:-

a)
$\frac{1}{1&space;-&space;cos\,\theta&space;}&space;+&space;\frac{1}{1&space;+&space;cos\;\theta&space;}&space;=&space;2\;cosec^2\,\theta$

$=&space;\frac{(1&space;+&space;cos\,\theta&space;)&space;+&space;(1&space;-&space;cos\,\theta&space;)}{1&space;-&space;cos^2\,\theta&space;}&space;=&space;\frac{2}{sin^2\,\theta&space;}&space;=&space;2cosec^2\;\theta$

b)

$(1&space;+&space;cot\,\theta&space;)(1&space;+&space;tan\,\theta&space;)&space;=&space;2&space;+&space;cosec\,\theta&space;\,sec\,\theta$

$(1&space;+&space;cot\,\theta&space;)(1&space;+&space;tan\,\theta&space;)&space;=&space;\left(1&space;+&space;\frac{1}{tan\,\theta&space;}&space;\right)\left(1&space;+&space;tan\,\theta&space;&space;\right)$

$=&space;\frac{\left(tan\,\theta&space;&space;+&space;1&space;\right)\left(1&space;+&space;tan\,\theta&space;&space;\right)}{tan\,\theta&space;}&space;=&space;\frac{tan^2\,\theta&space;&space;+&space;2tan\,\theta&space;&space;+&space;1}{tan\,\theta&space;}$

$=&space;\frac{2tan\,\theta&space;&space;+&space;sec^2\,\theta&space;}{tan\,\theta&space;}&space;=&space;2&space;+&space;\frac{sec^2\,\theta&space;}{tan\,\theta&space;}\;\;\;but\;\;\;(1&space;+&space;tan^2\,\theta&space;&space;=&space;sec^2\,\theta&space;)$

$\therefore\;\;\;&space;=&space;2&space;+&space;\frac{1}{cos^2\,\theta&space;}\times&space;\frac{cos\,\theta&space;}{sin\,\theta&space;}&space;=&space;2&space;+&space;cosec\,\theta&space;\;sec\,\theta$

c)

$cos^4\,\theta&space;&space;+&space;sin^4\,\theta&space;&space;=&space;1&space;-&space;2\;cos^2\,\theta\;&space;sin^2\,\theta$

$Since\;\;\;\;\;sin^2\,\theta&space;&space;+&space;cos^2\,\theta&space;&space;=&space;1$

$cos^4&space;=&space;(1&space;-&space;sin^2\,\theta&space;)^2&space;=&space;1&space;-&space;2\;sin^2\,\theta&space;&space;+&space;sin^4\,\theta$

$\therefore\;\;\;\;\;[cos^4\,\theta&space;&space;+&space;sin^4\,\theta&space;&space;=&space;1&space;-&space;2\;sin^2\,\theta&space;&space;+&space;2\;sin^4\,\theta$

$=&space;1&space;-&space;2\;sin^2\,\theta&space;&space;+&space;2\;sin^4\,\theta&space;&space;=&space;1&space;+&space;2sin^2\,\theta&space;(sin^2&space;-&space;1)$

$=&space;1&space;-&space;2cos^2\,\theta&space;\;sin^2\,\theta$

## The Graphs Of The Trigonometrical Ratios

In defining the ratios the following graph was used.

It can be seen that the value of $\inline&space;&space;sin\;\theta$ is given by the y-ordinate. Thus by drawing a circle of unit radius, the value of of the sine of any angle can be found. In this way the following graph was drawn. On the left hand circle $\inline&space;&space;\theta$ is measured from Ox in an anticlockwise direction whilst on the right hand graph $\inline&space;&space;\theta$ is measured along the x axis in the normal way.

$As\;\;\;\;\;\;\;Cos\,\theta&space;&space;=&space;sin\,(90^0&space;-&space;\theta&space;)$

the values taken by the cosine as the angle increases from 0 to 90 degrees will be the same as those taken by the sine as the angle decreases from 90 degrees to 0. The two graphs are identical in shape and magnitude but displaced by 90 degrees.

$eg.\;When\;\;\;\;\;\;\;\theta&space;&space;=&space;0,\;\;\;sin\,\theta&space;&space;=&space;0\;\;\;and\;\;\;cos\,\theta&space;&space;=&space;1$

To construct the graph of $\inline&space;&space;tan\;\theta$ is slightly more complicated. By inspection it can be seen that as:-

$tan\,\theta&space;&space;=&space;\frac{sin\;\theta&space;}{cos\,\theta&space;}$

The tan of the angle will be infinite whenever the cosine is zero.

To construct the graph, CX is drawn at unit length. The points $\inline&space;&space;P_1;P_2;P_3&space;etc$ are markedoff on the line XY and correspond to the various angles chosen for CP. If a point Q is plotted such that its abscissa on ON is equal to the number of degrees in the chosen angle XCP. This process is repeated for each value of the angle XCP.