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Trigonometric Ratios and Identities. - Best Maths Practice Material

# Trigonometric Ratios and Identities.

## Trigonometric Ratios and Identities

Trigonometric ratios and identities is one of very important topic for any type of school or competitive examination.

The objective of this chapter is, that after this we should be comfortable with angles and one of most important function trigonometric function

The flow of topics will be as

• Angles.
• Trigonometric identities.
• Compound angle formulas.
• Algebraic to product form.
• Product to algebraic form.
• Multiple angle and Half angle formulas.

## Angles.

Angle is one of most important part of trigonometry. Angle is measure of rotation of a given ray about its initial point. Where original ray before rotation is said to be initial ray and the final position of ray after rotation is called as terminal ray or final ray. The point about which ray is rotated is called as the vertex.

Generally initial ray is taken along the positive direction of x axis. Origin is taken to be the vertex of the angle.

The measure of angle is the amount by which the the ray is rotated, is positive when ray is rotated in anti-clockwise direction and angle measure is negative when rotated in clockwise direction.

### Angle measurement

Degree measure or British System.

In this system each right angle is divided into 90 equal parts is called as $1^0$ ( 1 degree ). Further $1^0$ is divided into 60 equal parts called as minute and each minuets is divided into 60 equal part called as seconds.

1 right angle = $90^0$

1 degree ($1^0$) = 60′      ( 60 minutes )

1 Minutes ( 1′) = 60″     ( 60 Seconds )

The angle subtended by the length of arc equal to the radius of the circle at center of the circle is called as one radian ( 1 radian ).

Conversion of degree into radians and vise versa.

$180^0$ = $\pi$ radians

Radian measure = $\cfrac{\pi}{180}$ x ( Degree measure )

Degree measure = $\cfrac{180}{\pi}$ x ( Radian measure )

#### Relation between Length of arc, radius of circle and angle subtended at center

Let O be the center of circle C with radius r, Let length of arc $\overset{\large\frown}{PQ}$ be $l$ and angle subtended at center of circle O is $\theta$.

We know that angle subtended at center of circle is directly proportional to length of arc.

When length of arc $l$ = r where r is radius of circle then angle subtended is 1 radian.

Hence $\cfrac{\text{(length of arc)}}{r}$ = $\cfrac{\theta}{1}$

Hence $\theta\,=\,\cfrac{l}{r}$

Example 1.

A wheel makes 30 revolutions per minute. Find the circular measure of the angle described by a spoke in $\cfrac{1}{2}$ second.

Solution 1.

Wheel is making 30 revolutions per minutes.

Hence angle subtended at center in 1 minute is = 30 x 2π = 60π

In 1 minutes we have 60 Seconds.

Hence in 60 Seconds the angle subtended at center = 60π radians.

Therefore in 1 second angle subtended at center = $\cfrac{\cancel{60}\pi}{\cancel{60}}$

Therefore angle subtended at center in $\cfrac{1}{2}$ = $\cfrac{1}{2}$(π) = $\cfrac{\pi}{2}$

Example 2.

In two circles arcs of equal length subtend angles of $60^0$ and $75^0$ at there centers. Find the ratio of there radii.

Solution 2.

Let Length of arc, radii and angle subtended at center of circles being $l_1,\,r_1,\,\theta_1$ and $l_2,\,r_2,\,\theta_2$ respectively.

Given $\theta_1\,=\,60^0$ and $\theta_2\,=\,75^0$

Since  $l_1\,=\,l_2$ ⇒ $r_1\theta_1$ = $r_2\theta_2$

Hence   $r_1\times 60^0$ = $r_2\times 75^0$

Therefore $\cfrac{r_1}{r_2}$ = $\cfrac{75}{60}$

Therefore $\cfrac{r_1}{r_2}$ = $\cfrac{5}{4}$

Hence radii are in ratio of 5:4.

### Relation between Interior angle and sides of regular polygon.

Let $A_1,A_2,A_3,…,A_n$ be the vertices of n sided regular polygon. Let O be the center of the polygon.

Considering the triangle $OA_1A_2$

Let $\angle OA_1A_2$ = $\angle OA_2A_1$ = $\theta$

Using sum of angles of triangle is $\pi$ radians. we get $\angle A_1OA_2$ = $\pi \,-\, 2\theta$

Hence Total angle at center of polygon = $n(\pi \,-\,2\theta)$ = $2\pi$

Hence   Sum of all internal angles.  2n$\theta$ = (n – 2)$\pi$   where 2$\theta$ is one of internal angle.

Hence each internal angle is (2$\theta$) = $\cfrac{(n-2)}{n}\pi$

Example 3.

Find each interior angle of a regular pentagon in circular measure.

Solution 3.

Given number of sides of pentagon (n) = 5

Hence each interior angle of pentagon = $\cfrac{(n\,-\,2)}{n}\pi$

Therefore for n = 5 each interior angle is = $\cfrac{5\,-\,2}{5}\pi$

Hence   Each interior angle of regular pentagon = $\cfrac{3}{5}\pi$

Example 4.

The number of sides in two regular polygons are in ratio 5 : 4 and difference between their internal angle is $9^0$, find the number of sides of of two polygons.

Solution 4.

Let number of sides of two polygons are 5n and 4n.

Given difference of internal angle is $9^0$

Therefore   $\cfrac{(5n\, -\, 2)}{5n}180^0$ – $\cfrac{(4n\,-\,2)}{4n}180^0$ = $9^0$

Therefore   4(5n – 2) – 5(4n – 2) = n

Hence   n = 2 therefor number of sides are 10 and 8.