Mathematical tools and vector Class - IX

CONCEPT OF TRIGONOMETRY:

Trigonometry is the part of mathematics where we learn about the angle. We know that in geometry the measured angle is restricted in between to and it is positive. But in trigonometry there is no limit of angle and it may be positive or negative.

Let us consider OA and OB are two straight lines initially coincide with each other. OB rotates about point O and creates angle keeping OA fixed. If OB rotates in anticlockwise direction then = is positive. If OB rotates in clockwise direction then = is negative.

CIRCULAR SYSTEM OF MEASURING ANGLE:

A particle rotates in a circular path of radius OA (=r) in anticlockwise direction. The particle is at point B creates an angle = .

The angle subtended at the centre of the circle by arc is the ratio of length of arc to the radius of the circle. = = .

In circular system the unit of angle is radian. The angle subtended at the centre of the circle by arc whose length is equal to the radius of the circle is called one radian.

RELATION BETWEEN RADIAN AND DEGREE:

A particle rotates in a circular path of radius r and completes one rotation. Then angle subtended at the centre is = .

In radian unit = = .

Therefore, radian =

Or, 1 radian = = =

PROVE = :

Let us consider a particle rotates in a circular path of radius R in anticlockwise direction starting from point A. The particle is at

point B creates an angle = . The length of arc AB = S.

So, = ——–(i)

Now we consider another point C on the circumference of the circle so that AC = R. Therefore the angle = 1 radian. So, = ——–(ii)

As angle subtended at the centre of a circle is directly proportional to its arc, then from equation (i) and (ii) we get, =

Or, = = [where is measured in circular system].

APPLICATION IN PHYSICS:

A particle rotates in a circular path of radius r in anticlockwise direction. If the linear speed of the particle is v then what is its angular speed?

Let us consider the particle starting from point A and comes at point B and the = . Length of arc AB = S and the time of travel is t. So, v = .

Angular speed of the particle is = = [as = ] Or, = () = .

TRIGONOMETRICAL RATIO OF POSITIVE ACUTE ANGLE:

Let us consider, OA and OB are two straight lines initially coincide with each other. OB rotates about point O and creates angle = keeping OA fixed. P is a point on OB and PQ is perpendicular on OA. Therefore, OQ is base, PQ is perpendicular and OP is the hypotenuse of OPQ.

sin = = , cos = = , tan = = ,

cosec = = , sec = = , cot = = .

Therefore, sincosec = 1, = 1, = 1

PROVE :

(i) + = 1

Using Pythagoras’s theorem in OPQ we get, + =

Or, + = 1

+ = 1

(ii) – = 1

Using Pythagoras’s theorem in OPQ we get, + =

Or, – =

Or, – = 1

– = 1.

(iii) – = 1

Using Pythagoras’s theorem in OPQ we get, PQ² + OQ² = OP²

Or, OP²– PQ² = OQ²

Or, = 1

sec² – tan² = 1

APPLICATION IN PHYSICS:

(i) An inclined plane AB of inclination is placed on ground at a height h. What is the length of AB?

Height of inclined plane BC = h. From ABC, = sin

Or, = sin AB = = hcosec.

(ii) A bob of mass m is suspended freely from the celling by a string. Now a horizontal force F is applied on the bob and the string creates angle with vertical. What is the tension acts on the string if the bob is at rest?

There are horizontal (H) and vertical (V) component of tension acts on the bob.

From diagram = so H = T and = so, V = T.

As the bob is at rest then, F = T and Mg = T

Now squaring and adding these two equations we get, + = ( + )

Or, = +

T = .

TRIGONOMETRICAL RATIO OF SOME STANDARD ANGLES:

	or	or	or	or	or
sin	0				1
cos	1				0
tan	0		1
cosec		2			1
sec	1			2
cot			1		0

APPLICATION IN PHYSICS:

A block is moved point A to B (AB = 10m) by applying a force F = 10 N at an angle with AB. What is the work done required?

Work (W) = force (F) displacement.

The component of force along AB is H and = cos

Therefore, H = Fcos = 10cos = = 5

W = 5 10 = 50 J.

TRIGONOMETRICAL RATIO OF COMPLEMENTARY ANGLE:

If the sum of two angles is , then those two angles are called complementary angles. In figure OAB is a right-angled triangle. AOB = and OBA = and + = and = ( – ).

So, and are the complementary angles. From , OAB, sin = and cos = . Therefore, sin = cos = cos( – ). Similarly,

sin( –) = cos	sec( –) = cosec
tan( –) = cot	cot( –) = tan
cosec( –) = sec

TRIGONOMETRICAL RATIO OF ASSOCIATED ANGLE:

Let us consider, XOX^/ and are two mutually perpendicular axis where O is the origin. OA is a straight line which rotates about O in anticlockwise direction and creates an angle (acute angle). P is a point on OA and PQ is perpendicular on OX axis. From diagram OP, OQ and PQ are positive.

So, sin = is positive, cos = is positive and tan = is positive. Similarly , cosec, sec and cot also positive.

Therefore in 1^st quadrant all trigonometrical ratio are positive.

When OA is in 2^nd quadrant then from diagram OP and PQ are positive but OQ is negative and + = . So, sin = is positive, cos = is negative and tan = is negative. Similarly cosec is positive but sec and cot are negative.

When OA is in 3rd quadrant then from diagram OP is positive but OQ and PQ are negative and + = .

So, sin = is negative, cos = is negative and tan = is positive. Similarly cosec and sec are negative but cot is positive.

When OA is in 4th quadrant then from diagram OP and OQ are positive but PQ is negative and + = .

So, sin = is negative, cos = is positive and

tan = is negative.

Similarly cosec and cot are negative but sec is positive.

If is positive acute angle associated with the angle (n ) [where n is any positive integer (1, 2, 3, —–)], then to find the value of any trigonometrical ratio of this angle at first identify the quadrant in which the angle (n ) is located and use positive or negative sign with respect to the quadrant.

When n is even	When n is odd
sin(n ) = sin	sin(n ) = cos
cos(n ) = cos	cos(n ) = sin
tan(n ) = tan	tan(n ) = cot
cosec(n ) = cosec	cosec(n ) = sec
sec(n ) = sec	sec(n ) = cosec
cot(n ) = cot	cot(n ) = tan

EXAMPLE:

sin = sin(4 – ) = – sin = – .

sin = sin(3 + ) = – cos = – .

cos = cos(4 + ) = + cos = + .

TRIGONOMETRICAL RATIO WHEN ANGLE IS VERY SMALL:

When is very small (less than ) then sin = , cos = 1 and tan = . For example, = = 0.017453 and

sin = 0.017452.

PARALLAX METHOD:

If you hold a pen in front of your eye and look the tip of the pen by closing the right eye and then the left eye keeping right eye open, then you observe that the position of the tip of the pen changes. This relative shift of the tip of the pen with respect to background is called parallax. The distance between two points of observations (the distance between your eyes) is called the basis. The angle subtended by basis at the object is called parallactic angle.

If a planet of diameter d is at a distance D (D d) and is the parallactic angle, then is very small. From diagram we can write, tan = or, = .