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

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.

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

Trigonometric Ratios:- cos

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

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.

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 ’ 0 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.

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

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 .

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.

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 ’ 0 M M ’ A Y ’

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 2) 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

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 .

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 = 1/0 . Y ‘ ….. is no t defined 1 O

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

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.

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

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 = √ 32 + 42 = 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

Trigonometric Identities : - There are three Fundamental Trigonometric Identities viz : For any θ Є R , 1) sin2θ + cos2θ = 1 2) 1 + tan2θ = sec2θ 3) 1 + 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 ) .

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

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

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

3. Dividing both sides of equation ( i ) by y2 , if y ≠ 0 we get , x 2 /y2 + y 2 / y 2 = r 2 / y 2 ( x /y ) 2 + 1 = ( r / y )2 cot 2 θ + 1 = cosec 2 θ 1 + cot 2 θ = cosec 2 θ Also we can write : a) cosec 2 θ - 1 = cot 2 θ b) cosec 2 θ - cot 2 θ = 1

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 + 2 = 0 . ( 3 tan x - 2 ) ( tan x - 1 ) = 0 . tan x = 1 or tan x = 2/3

EXAMPLE :- If θ = - 60º then find the value of all trigonometric ratios . Sol . :- If θ = - 60º then 1) Sin θ = sin ( - 60º ) = - sin 60º= -√3/2 2) cos θ = cos(-60º)=cos 60º= (1/2) 3) tanθ = (tan-60º)= - tan 60º= - √3 4) cosec θ = cosec (-60º) = - cosec 60º= - 2/√3 5) sec θ = sec ( - 60º ) = sec 60º= 2 6) cotθ = (cot-60º)= - cot 60º = -1/√3