Basic Maths Section [B] Trigonometry

SECTION [B] : TRIGONOMETRY ANGLE R ad iu s it is measure of change in direction. s r Arc  s  Angle () = Radius r  O arc r Radius Angels measured in anticlockwise and clockwise direction are usually taken positive and negative respectively. System of measurement of an angle [A] Sexagesimal system : In this system, angle is measured in degrees. In this system , 1 right angle = 90°, 1° = 60' (arc minutes), 1' = 60'' (arc seconds) [B] Circular system : In this system, angle is measured in radian. if arc = radius then =1 rad Relation between degrees and radian 2 rad = 360°  rad = 180°  1 rad = To convert from degree to radian multiply by  180 To convert from radian to degree multiply by 180  180 = 57.3°  Ex. A circular arc of length  cm. Find angle subtended by it at the centre in radian and degree. Sol.  Ex. When a clock shows 4 o'clock, how much angle do its minute and hour needles make? s  cm    rad=3 0° r 6 cm 6 (A) 120° Sol. 6cm (B) cm 6cm  rad 3 (C) 2 rad 3 (D) 160° (AC) 11 12 1 10 2 9 3 8 4 7 6 5 From diagram angle   4  3 0   1 20   2 rad 3 Ex. Sol. The moon’s distance from the earth is 360000 km and its diameter subtends an angle of 42' at the eye of the observer. The diameter of the moon in kilometers is (A) 4400 (B) 1000 (C) 3600 (D) 8800 (A) Here angle is very small so diameter  arc 0 1  1  7    42   42    42    rad 60  60 180 1800  Diameter = R  360000  7 22   4400 km 1800 7 Trigonometric Ratios (T-ratios) Following ratios of the sides of a right angled triangle are known as trigonometrical ratios. P H cos   B H tan   P B cosec  sec   1 H  cos  B cot   Perpendicular (P) sin   1 H  sin  P 1 B  tan  P Base(B) Trigonometric Identities 2 2 P B Divide by H2,       1  sin 2   cos 2   1 H  H  In figure , P2 + B2 = H2 2 P H  Divide by B2,    1    B B 2 2  1  tan 2   sec 2  2 B H  Divide by P2,  1        1  cot 2   cosec2 P  P  Commonly Used Values of Trigonometric Functions A n gle( ) 0° 30° 37° sin  0 1 2 3 5 cos 1 4 5 tan  0 3 2 1 3 45° 1 3 4 2 1 2 1 53° 60° 4 5 3 2 1 2 3 5 4 3 3 1 0  90° Four Quadrants and ASTC Rule In first quadrant, all trigonometric ratios are positive. In second quadrant, only sin and cosec are positive. In third quadrant, only tan and cot are positive. In fourth quadrant, only cos and sec are positive 90° II quadrant Sin Ist quadrant All Tan IIIrd quadrant Cos IVth quadrant nd 180° Remember as ‘Add Sugar To Coffee’ or ‘After School To College’. Trigonometrical Ratios of General Angles (Reduction Formulae) (i) Trigonometric function of an angle 2n +  where n=0, 1, 2, 3,.... will be remains same. sin(2n +) = sin cos(2n+) = cos tan(2n+)=tan 270° 0° 360° (ii)  n  Trigonometric function of an angle     will remains same if n is even and sign of trigonometric function will  2  be according to value of that function in quadrant. sin(–) = + sin cos() = –cos sin()= –sin cos()= –cos sin()= –sin cos()= +cos (iii) tan()= –tan tan()= +tan tan()= –tan  n    will be changed into co-function if n is odd and sign of trigonometric Trigonometric function of an angle   2 function will be according to value of that function in quadrant.   sin      cos  2    cos      sin  2    tan      cot  2    sin      cos  2    cos      sin  2    tan      cot  2  (iv) Trigonometric function of an angle – (negative angles) sin(–) = – sin cos(–) = + cos tan(–) = – tan Ex. The two shorter sides of right angled triangle are 5 cm and 12 cm. Let  denote the angle opposite to the 5 cm side. Find sin, cos and tan. Sol. sin   P 5 cm 5   H 1 3cm 1 3 cos   B 12 cm 12   H 13 cm 13 tan   P 5 cm 5   B 12 cm 1 2 Ex. Find x, y and perimeter of the triangle Sol. y 4  sin 53   y  4 cm 5 5 and 5cm 53° y x 3  cos 53    x  3 cm 5 5 x Perimeter of the triangle = x + y + 5 =3 + 4 + 5 = 12 cm Ex. Find the value of : (i) sin30° + cos60° (v) cos 405° (ix) tan 135° (xiii) cos(–60°) Sol. (i) sin 30°+cos 60° = (ii) sin 0° – cos 0° (iv) tan 420° (x) sin (330°) (xiv) tan(–45°) 1 1 + =1 2 2 (iii) tan 45°–tan 37° = 1  3 1  4 4 (iii) tan 45°–tan 37° (viii) sin 150° (xi) cos 300° (xvi) sin(–150°) (iv) sin 390° (viii) cos 120° (xii) sin(–30°) (ii) sin 0°–cos0°=0–1 = –1 (iv) sin 390°=sin(360°+30°) = sin 30°= 1 2 (v) cos 405° = cos(360°+45°)=cos45° = (vii) sin 150° = sin (90°+60°) = cos 60° = 1 (vi) tan 420° = tan (360°+60°) = tan 60°= 2 1 1 or sin 150°=sin(180°–30°)=sin30°= 2 2 (viii) cos 120°=cos(180°–60°)= – cos60° = – (x) sin 330° = sin (360°–30°) = – sin 30° = – 1 2 1 2 1 2 (xiv) tan (–45°) = –tan45° = –1 (xii) sin (–30°)= –sin 30° = – (ix) tan 135°=tan(180°–45°) = –tan 45°= –1 (xi) cos 300° = cos (360°–60°) = cos 60° = Sol. 1 2 1 2 (xv) sin (–150°) = –sin (150°) = – sin (180°–30°) = – sin 30° (xiii) cos (–60°) = + cos 60° = =– Ex. 1 2 1 2 The values of sin1, cos22 and tan3 are given as ½ , – ½ and 3 (not in order), for some angles 1, 2 and 3. Choose incorrect statement. (A) The value of tan3 could be –½ (B) The value of sin1 can not be 3. (C) The value of cos22 can't be –½ (D) The value of cos22 could be 3. (D) 1  sin 1  1 , 0  cos2 2  1 ,   tan 3   Addition/Subtraction Formulae for Trigonometrical Ratios (i) sin (A+B)=sinA cosB + cosA sinB (iii) cos(A+B)=cosA cosB –sinA sinB Ex. Sol. (ii) sin(A–B)=sinA cosB–cosA sinB (iv) cos(A–B)=cosA cosB + sinA sinB By using above basic addition/ subtraction formulae, prove that tan A  tan B tan A  tan B (i) tan (A+B) = (ii) tan (A–B) = 1  tan A tan B 1  tan A tan B 2 tan  (iv) cos2 = cos2–sin2 = 1–2sin2 = 2cos2–1 (v) tan2 = 1  tan 2  sin  A  B  sin A cos B  cos A sin B (i) tan(A+B) = cos A  B  cos A cos B  sin A sin B   (iii) sin2 = 2sin cos  sin A sin B  cos A cos B     cos A cos B   tan A  tan B = sin A sin B  1  tan A tan B  cos A cos B 1   cos A cos B   (ii) sin  A  B  sin A cos B  cos A sin B tan (A–B) = cos A  B  cos A cos B  sin A sin B    sin A sin B  cos A cos B    cos A cos B   tan A  tan B   sin A sin B  1  tan A tan B cos A cos B 1   cos A cos B  (iii) (iv) (v) Ex. sin2 = sin(+) = sin cos + cos sin = 2 sin cos cos2 = cos(+) = cos cos–sinsin = cos2–sin2 = 1–sin2–sin2 = 1–2sin2 = 1–2(1–cos2) = 2 cos2 –1 tan   tan  2 tan   tan2 = tan(+) = 1  tan  tan  1  tan 2  Find the value of (i) sin 74° (ii) cos 106° (iii) sin 15° (iv) cos 75° Sol.  3  4  24 (i) sin 74° = sin (2× 37°) = 2 sin 37 cos 37° =2      5  5  25 2 2 (ii) 7  3   4  9  16  cos 106° = cos (2×53°) = cos 53° sin 53° =       5 5 25 25     (iii) sin 15° = sin (45°–30°) = sin 45° cos 30° – cos 45° sin 30° = (iv) cos 75° = cos (45°+30°) = cos 45° cos30° – sin 45° sin 30° = 2 2 Small Angle Approximation If is small (say < 5°) then sin   , cos  Note : here  must be in radian. 2 2  3 3    2 2 2 2 2  3 3    2 2 2 2 2  1 and tan   Ex. Find the approximate values of (i) sin 1° (ii) tan 2° (iii) cos 1°. Sol.       (i) sin1  sin 1   sin 180  180 180       (ii) tan 2  tan  2   tan  180  90 90      1 (iii)  cos1  cos 1   cos 180  180   Maximum and Minimum Values of Some useful Trigonometric Functions Ex. Find maximum and minimum values of y : (i) y = 2 sinx Sol. 2 2 2 2 (iii)  a  b  a cos   b sin   a  b (ii) 1  cos   1 (i) 1  sin   1 (ii) y = 4–cosx (iii) y = 3sinx + 4 cosx (i) ymax =2(1)=2 and ymin=2(–1)=–2 (ii) ymax = 4 – (–1)=4+1=5 and ymin = 4–(1)=3 (iii) ymax = Ex.  2 3  42 =5 and ymin = – 3  4 = – 5 A ball is projected with speed u at an angle to the horizontal. The range R of the projectile is given by R u 2 sin 2 g for which value of  will the range be maximum for a given speed of projection? (Here g = constant) (A) Sol.  rad 2 (B)  rad 4 (C)  rad 3 (D)  rad 6 (B) As sin 2  1 so range will be maximum if sin 2 =1. Therefore 2 =      rad. 2 4 Ex. The position of a particle moving along x-axis varies with time t according to equation x = 3 sint – cost where  is constants. Find the region in which the particle is confined. Sol. Q x  3 sin t  cos t  xmax   3    1  2 and xmin  2  3    1  2 Thus, the particle is confined in the region 2  x  2 2