1. P.J.Mhatre Vidyalay Nawade
Name: Sou.Devkar S.S
Std: IX th
Sub: Algebra

2. TRIGONOMETRY Introduction :-
The word trigonon means a triangle and the word metron means a measures .
Hence, trigonometry means the science of measuring triangles. I N broader sense it is that branch of mathematics which deals with the measurement of of the sides and angles of a triangle.

3. Abbreviations for Trigonometric ratios:-
Sine of angle θ = sin θ
Cosine of angle θ = cos θ
Tangent of angle θ = tan θ
Cosecant of angle θ = csc θ
Secant of angle θ = sec θ
Cotangent of angle θ = cot θ

4. Trigonometric Ratios:-
cos

5. Sinθ increases from 0 to 1 where as Cosθ decreases from 1 to 0 .

6. Do you observe any relation between the pair Sin30º and Cos60º, tan45º
And Cot45º, Also sec60º and cosec30º ?
We get identities for complementary angles.
Sin30º = Cos(90º - 30º) = Cos60º
tan45º = Cot(90º - 45º) = Cot45º
Sec60º = Cosec(90º - 60º) = Cosec30º
So you all know the identities for complementary angles.

7. In rectangular co-ordinate system take a point P (x,y) where x Є R .
Draw seg PM which is perpendicular to X-axis. The number corresponding to the point M on X-axis is called x-coordinate or abscissa. Draw seg PN which is perpendicular to the point M on Y-axis .the number corresponding to the point N on the y-axis is called y-coordinate or ordinate. Y
x P(x,y) Point P is (x,y) and the origin 0 is (0,0)
N By distance formula
y OP = x2+y2 .
Also PM = y and PN = x
X ’ M X In general, we say that x-co-ordinate or
y-co-ordinate of a point P is positive or
Y ’ negative depending on the quadrant P lies.

8. Standard Angle : - Measure Of Standard Angle : - Y
In rectangular co-ordinate system a directed angle with its vertex at the origin O and initial ray along positive X-axis is called Standard Angle or Angle in Standard Position .
Vertex of angle in standard position :- In rectangular coordinate system the origin O is the vertex of angle in standard position.
Measure Of Standard Angle : Y
For <AOP, ray OA is the initial arm, P
Ray OP is the terminal arm , and θ is
the amount of rotation X’ θ X
For <AOQ, OA is the initial arm, OQ is O α A
the terminal arm, and α is the amount
of rotation Y’ Q

9. Angle in Quadrant :- Y A P3 Y’
X’ O X A P3 Y’
In the fig. the terminal arm OP3 is ÌÌÌ quadrant so we say that , <AOP3 is in ÌÌÌ quadrant .
Here the measure of the <AOP3 lies between 180º and 270º or -180ºand -90º .

10. Quadrant Angle : -
A Directed Angle in standard position whose terminal arm lies along co-ordinate axes is called Quadrant angle. Y B X
O A D
Y ’
In fig < AOB, <AOC, <AOD and <AOA are all Quadrantal angles.

11. Trigonometric ratios in terms of co-ordinates of a point :-Y
B Let <AOB = θ be in standard position
whose initial arm is OA and terminal
N ’ …………………… arm is OB. Let P(x,y) be a point on
P( x ’, y ’) the terminal arm OB such that OP= r
so x2+y2 = r2 (by distance formula .
N ……………
P(x,y) Trigonometric ratios in terms of
r co-ordinates of point P(x,y)
θ X
X ’ M M ’ A
Y ’

12. For our convenience if we take
point P on a standard unit circle whose
centre is at the origin and radius r = 1 .
Then we get the trigonometric
Ratios as follows :
1) sin θ = y ) cos θ = x
3) tan θ = y/x , where x ≠ 0
4) cosec θ = 1/y , where y ≠ 0
5) sec θ = 1/x , where x ≠ 0
6) cot θ = x/y , where y ≠ 0

13. Signs of trigonometric ratios in different quadrant : -
If x is positive , cosine is positive .
If x is negative , cosine is negative .
If y is positive , sine is positive .
If y is negative , sine is negative .

14. Trigonometric Ratios of 0º : - Y
sin 0º = y = 0 , cosec 0º = 1/y = 1/0 .
….. is no t defined P(1,0)
X ’ A X
cos 0º = x = 0 , sec 0º = 1/x = 1/1 = 1 .
tan0 º = y/x = 0/1 = 0 , cot 0º = x/y = 1/ Y ‘
….. is no t defined 1 O

15. Trigonometric Ratios of 90º : - Y B P(0,1)
sin 90º = y = 1 , cosec 90º = 1/y = 1/1 = 1 .
X ’ A X
cos 90º = x = 0 , sec 90º = 1/x = 1/0
….. is no t defined .
tan90 º = y/x = 1/0 ….. is no t defined Y ‘
cot 90º = x/y = 0/1
90º O

16. Trigonometric ratios of negative angles : -
For any angles having measure θ .
Example.
sin (- θ ) = - sin θ
Proof :-
Consider a standard unit circle, which
cuts X-axis at the point A.
Rotate the ray OA around O in anti-clockwise direction making an angle θ , intersecting circle at the point P (x , y) , so that ray OP is the terminal arm.
Now, rotating OA around O in anti- clockwise direction making an angle θ , intersecting the circle at point P (x , y), so that ray OQ is the terminal arm.

17. So .. <AOP = θ and <AOQ = - θ
The points P and Q are mirror images of each others.
x-co-ordinate of Q = x-co-ordinate of P So x ’ = x .
y-co-ordinate of Q = -(y co-ordinate of P) So y ’ = - y .
So P ≡ (x , y) and Q ≡ (x,-y)
By definition of Trigonometric ratios in standard position ,
sin θ = y and sin( - θ ) = - y
So sin ( - θ ) = - sin θ

18. Example :–
Find the trigonometric ratios in standard position whose terminal arm passes through ( 3 , 4 ) .
Solution :-
let the terminal arm passes through ( 3 , 4 ) So X = 3 and Y = 4 .
r = √ x2 + y2 = √ = 5
and let the angle having measure θ
So.
sin θ = y/r = 4/5 , cosec θ = r/y = 5/4
cos θ = x/r = 3/5 , sec θ = r/x = 5/3
tan θ = y/x = 4/3 , cot θ = x/y = 3/4

19. Trigonometric Identities : -
There are three Fundamental Trigonometric
Identities viz : For any θ Є R ,
1) sin2θ + cos2θ = 1
2) tan2θ = sec2θ
3) cot2θ = cosec2θ
Proof :-
let us consider a standard circle of radius r. Let the circle intersect X-axis at the point A . Let the initial arm rotate in an anti – clockwise direction making an angle θ .
The terminal arm of angle θ intersects the circle at the point P (x,y)
such that x , y ≠ 0 . OP = r ( radius ) .

20. By definition of Trigonometric ratios in standard position we get ,
Since OP = r
So √ x2 + y2 = r …….( by distance formula )
or x2 + y = r2 …….( i )

21. 1. Dividing both sides of equation ( i ) by r2 , we get
x 2 /r y 2 / r = r 2 / r 2
( x/r ) ( y/ r ) =
cos 2 θ + sin 2 θ = 1
sin 2 θ + cos 2 θ = 1
Also we get :
a) cos 2 θ = sin 2 θ
b) sin 2 θ = cos 2 θ

22. 2. Dividing both sides of equation ( i ) by x2 , if x ≠ 0 we get ,
x 2 /x y 2 / x = r 2 / x 2
( y / x ) = ( r / x )2
1 + tan 2 θ = sec 2 θ
Also we can write :
a) sec 2 θ = tan 2 θ
b) sec 2 θ - tan 2 θ =

23. 3. Dividing both sides of equation ( i ) by y2 , if y ≠ 0 we get ,
x 2 /y y 2 / y = r 2 / y 2
( x /y ) = ( r / y )2
cot 2 θ = cosec 2 θ
cot 2 θ = cosec 2 θ
Also we can write :
a) cosec 2 θ = cot 2 θ
b) cosec 2 θ - cot 2 θ =

24. Example :–
Find the possible values of tan x ,
if cos 2 x + 5 sin x . cos x = 3 .
Solution : Given , cos 2 x + 5 sin x . cos x = 3 .
Dividing both sides by cos 2 x , we
get,
1 + 5 tan x = 3 sec 2 x .
1 + 5 tan x = 3 (1 + tan 2 x) .
3 tan 2 x - 5 tan x = 0 .
( 3 tan x - 2 ) ( tan x ) = 0 .
tan x = 1 or tan x = 2/3

25. EXAMPLE :- If θ = - 60º then find the value of all trigonometric ratios . Sol . :- If θ = - 60º then 1) Sin θ = sin ( - 60º ) = - sin 60º= -√3/ ) cos θ = cos(-60º)=cos 60º= (1/2) ) tanθ = (tan-60º)= - tan 60º= - √3 4) cosec θ = cosec (-60º) = - cosec 60º= - 2/√3 5) sec θ = sec ( - 60º ) = sec 60º= ) cotθ = (cot-60º)= - cot 60º = -1/√3