Trigonometry Notes

TRIGONOMETRY Page # 1 TRIGONOMETRY SYSTEM OF MEASUREMENT OF ANGLE There are three system for measuring angles. Sexagesimal or English system Centesimal or French system Circular system Sexagesimal system : The principal unit in this system is degree (º). One right angle is divided into 90 equal parts and each part is called one degree (1º ). One degree is divided into 60 equal parts and each part is called one minute. Minute is denoted by (1'). One minute is equally divided into 60 equal parts and each part is called one second (1''). In Mathematical form : One right angle = 90º 1º = 60 ' 1' = 60 '' Centesimal system : The principal unit in system is grade and is denoted by ( g ). One right angle is divided into 100 equal parts, called grades, and each grade is subdivided into 100 minutes, and each minutes into 100 seconds. In Mathematical Form : One right angle = 100g g 1 = 100 ' 1' = 100' Relation between sexagesimal and Centesimal systems : One right angle = 90º (degree system) ...(1) One right angle = 100g (grade system) ...(2) by (1) and (2), 90º = 100g measure of AOB is 1 radian (1C). Relation between systems of measurement of angles : D G 2C   90 100  Fundamental Trigonometrical Identities : (a) sin = 1 cos ec (b) cos = 1 sec  (c) cot = 1 cos   tan  sin  (d) 1 + tan2 = sec2 or, sec2 – tan2 = 1 (sec – tan) = 1 (sec   tan ) (e) sin2 + cos2 = 1 (f) 1 + cot2 = cosec2 (cosec – cot) = 1 cos ec  cot  Signs of the trigonometrical ratios or functions: D G  90 100 then we can say, or, 1º = FG 100 IJ H 90 K g Circular system : One radian, written as 1C, is the measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. Consider a circle of radius r having centre at O. Let A be a point on the circle. Now cut off an arc AB whose length is equal to the radius r of the circle. Then by the definition the A crude aid to memorise the signs of trigonometrical ratio in different quadrant. TRIGONOMETRY Page # 2 cosec = GRAPH OF DIFFERENT TRIGONOMETRICAL RATIOS Y Y –3/2 –/2 sin  = X' –2 – y =1 /2 3/2  2 X O y = –1 (–3/2, 1) X' O (–/2, –1) Y' Y cos  = X' –3/2 O – –/2 /2 X (3/2, –1) y = –1 Y' y=1  X 3/2 y = –1 Y' TRIGONOMETRICAL RATIO OF ALLIED ANGLES sin (90º – ) = cos cos (90º – ) = sin tan (90º – ) = cot cot (90º – ) = tan cosec(90º – ) = sec sec (90º – ) = cosec Y X O Y' sin (90° + ) = cos  cos (90° + ) = – sin tan (90° + ) = – cot cot (90° + ) = – tan  sec(90° + ) = – cosec  cosec (90° + ) = sec  Periodic Function : All the trigonometric function are periodic function. They repeat their value after a certain period sin (2n +  ) = sin cos (2n +  ) = cos tan (n +  ) = tan  SUM & DIFFERENCE FORMULAE (a) sin (A + B) = sin A cos B + cos A sin B (b) sin(A – B) = sinA cosB – cosA sinB (c) cos (A + B) = cos A cos B – sin A sin B (d) cos (A – B) = cos A cos B + sin A sin B Y y=1 (0,1) – y=1 (/2, 1) X O (e) tan (A + B) = tan A  tan B 1  tan A tan B (f) tan (A – B) = tan A  tan B 1  tan A tan B (g) cot (A + B) = cot A cot B  1 cot B  cot A (h) cot (A – B) = cot A cot B  1 cot B  cot A y = –1 Y' 1 TRIGONOMETRY Some More Results : *(a) sin (A + B).sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A *(b) cos (A + B).cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A (c) sin (A + B + C) = sin A cos B cos C + cosA sin B sin C + cos A cos B sin C – sin A sin B sin C (d) cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sinB cosC (e) tan (A + B + C) tan A  tan B  tan C  tan A tan B tan C 1  tan A tan B  tan B tan C  tan C tan A (Note : * Important) = FORMULAE FOR PRODUCT INTO SUM OR DIFFERENCE CONVERSION We know that, sin A cos B + cos A sin B = sin (A+B) .......(i) sin A cos B – cos A sin B = sin (A – B) ......(ii) cos A cos B – sin A sin B = cos (A+B) .....(iii) cos A cos B + sin A sin B = cos (A – B) .....(iv) Adding (i) and (ii), 2 sin A cos B = sin (A + B) + sin (A – B) Subtracting (ii) from (i), 2 cos A sin B = sin (A + B) – sin (A – B) Adding (iii) and (iv), 2 cosA cos B = cos (A + B) + cos (A – B) Subtraction (iii) from (iv). 2 sin A sin B = cos (A – B) – cos (A + B) Formulae : (a) 2 sin A cos B = sin (A + B) + sin (A – B) (b) 2 cos A sin B = sin (A + B) – sin (A – B) (c) 2 cos A cos B = cos (A + B) + cos (A – B) (d) 2 sin A sin B = cos (A – B) – cos (A + B) FORMULAE FOR SUM OR DIFFERENCE INTO PRODUCT CONVERSION We know that, sin (A + B) + sin(A – B) = 2 sin A cos B ......(i) Page # 3 Let A + B = C and A – B = D then A = C D C D and B = 2 2 Substituting in (i), (a) sin C + sin D = 2 sin FG C  D IJ .cos FG C  D IJ H 2 K H 2 K similarly other formula are, (b) sin C – sin D = 2 cos FG C  D IJ .sin FG C  D IJ H 2 K H 2 K (c) cos C + cos D FG C  D IJ .cos FG C  D IJ H 2 K H 2 K F C  D IJ .sin FG D  C IJ – cos D = 2 sin G H 2 K H 2 K = 2 cos (d)cos C TRIGONOMETRICAL RATIOS OF MULTIPLE ANGLES (i) sin 2 = 2 sin  cos  = 2 tan  1  tan2  (ii) cos 2 = cos2  – sin2  = 2 cos2  – 1 = 1 – 2 sin2  = (iii) tan 2 = 1  tan 2  1  tan 2  2 tan  1  tan2  (iv) sin 3 = 3 sin  – 4 sin3  (v) cos 3 = 4 cos3  – 3 cos  (vi) tan 3 = (vii) sin /2 = (viii) cos /2 = (ix) tan /2 = = 3 tan   tan 3  1  3 tan 2  1  cos  2 1  cos  2 1  cos  1  cos  1 cos  sin  = sin  1 cos  THE GREATEST AND LEAST VALUE OF THE EXPRESSION [a sin + b cos] Let a = r cos ...(1) and b = r sin  ...(2) Squaring and adding (1) and (2) then a2 + b2 = r2 TRIGONOMETRY Page # 4 or, r  a 2  b 2  General solution of the equation cos  = cos  : a sin + b cos = r (sin cos + cos sin)  = 2n –  ; n  I = r sin(+ )  = 2n +  ; n  I and But –1  sin   1 for the general solution of cos  = cos , so –1  sin( + )  1 combine these two result which gives. then –r  r sin( + )  r  = 2n ±  , n  I hence, General solution of the equation  a 2  b2  a sin + b cos  greatest a2  b2 then the  = n +  ; and least values of a sin + b cos are respectively 2 a b 2 and 2  a b tan  = tan  : n  I GENERAL SOLUTION OF SQUARE OF THE 2 TRIGONOMETRICAL EQUATIONS GENERAL SOLUTION OF STANDARD  = n ±  TRIGONOMETRICAL EQUATIONS General solution of cos2 = cos2 Since Trigonometrical functions are periodic  = n ±  functions, therefore, solutions of Trigonometrical equations can be generalised with the help of periodicity of Trigonometrical functions. The solution consisting of all possible solutions of a Trigonometrical equation is called its general solution. ; n  I ; n  I General solution of tan2 = tan2 :   = n ±  ; n  I SOME USEFUL RESULTS In a triangle ABC, p , h sin  = General Solution of the equation sin  = 0 : cos  = A b h when sin  = 0 is  = n : n  I i.e. n = 0, ± 1, ± 2 ......... h General solution of cos  = 0 :  when cos = 0  = (2n + 1) /2 , b C In any triangle ABC, n  I. i.e. n = 0, ± 1, ± 2 ........ General solution of tan  = 0 : when tan = 0 it same as of sin = 0 so that, general solution of tan   = n ; = 0 is n  I or cosine formula b2  c 2  a2 cos A = 2 bc A i.e., =0 does not exist because sec and cosec can c never be equal to 0. sin  = sin  :  = n + (–1)n  ; n  I B a b c = = [By sine rule] sin A sin B sin C Note : General solution of sec = 0 and cosec General solution of the equation p B b a C