TOPICS
Unit-3(Examples)

Example-1 :-  Convert 40° 20′ into radian measure.

Solution :-
  We know that 180° = π radian.
  Hence,
  40° 20′ = 40 + 1/3 degree 
          = π/180 x 121/3 radian 
          = 121π/540 radian
  Therefore, 40° 20′ = 121π/540 radian
   

Example-2 :-  Convert 6 radians into degree measure.

Solution :-
  We know that π radian = 180°.
  Hence, 
  6 radians = 180/π x 6 degree 
            = (1080 x 7)/22 degree 
            = 343° + 7/11 degree 
            = 343° + (7 x 60)/11 minute    [1° = 60']
            = 343° + 420/11 minute
            = 343° + 38' + 2/11 minute     [1' = 60"]
            = 343° + 38' + 10.9" 
  6 radians = 343° 38' 11" approximately.
   

Example-3 :- Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. (use π = 22/7)

Solution :-
   Given that l = 37.4 cm and θ = 60° = 60π/180 radian = π/3
   By using formula,
   r = l/θ 
     = (37.4 x 3)/π 
     = (112.2 x 7)/22
     = 785.4/22 
     = 35.7 cm
    

Example-4 :-  The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14).

Solution :-
  In 60 minutes, the minute hand of a watch completes one revolution.
  Therefore,in 40 minutes, the minute hand turns through 2/3 of a revolution.
  Therefore, θ = 2/3 x 360° = 4π/3 radian and r = 1.5 cm
  Hence, the required distance travelled is given 
  By formula,
  l = r θ 
    = 1.5 x 4π/3 
    = 2π 
    = 2 x 3.14
    = 6.28 cm
   

Example-5 :-  If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii.

Solution :-
  Let r₁  and r₂ be the radii of the two circles. Given that 
  θ₁  = 65° = π/180 x 65 = 13π/36 radian
  θ₂ = 110° = π/180 x 110 = 22π/36 radian
  Let l be the length of each of the arc. 
  Then l =  r₁ θ₁  =  r₂θ₂, which gives
  13π/36 x r₁  = 22π/36 x r₂ i.e.
  r₁ /r₂ = 22/13
  Hence, r₁  : r₂ = 22 : 13
   

Example-6 :-  If cos x = –3/5 , x lies in the third quadrant, find the values of other five trigonometric functions.

Solution :-
  Given that cos x = –3/5
(i) sec x = 1/cos x = -5/3
(ii) sin x = √1 - cos²x
           = √1 - (-3/5)²x
           = √1 - 9/25
           = √16/25
           = ±4/5
  Since x lies in second quadrant, sin x will be negative. Therefore
  sin x = -4/5
(iii) cosec x = 1/sin x = -5/4
(iv) tan x = sin x/cos x 
           = -4/5 ÷ (-3)/5 
           = -4/5 x (-5)/3
           = 4/3
(v) cot x = 1/tan x = 3/4
    

Example-7 :-  If cot x = –5/12, x lies in second quadrant, find the values of other five trigonometric functions.

Solution :-
  Given that cot x = –5/12
(i) tan x = 1/cot x = -12/5
(ii) sec x = √1 + tan²x
           = √1 + (-12/5)²x
           = √1 + 144/25
           = √169/25
           = ±13/5
  Since x lies in second quadrant, sec x will be negative. Therefore
  sec x = -13/5
(iii) cos x = 1/sec x = -5/13 
(iv) sin x = √1 - cos²x
           = √1 - (-5/13)²x
           = √1 - 25/169
           = √144/169
           = ±12/13
  Since x lies in second quadrant, sin x will be positive. Therefore
  sin x = 12/13
(v) cosec x = 1/sin x = 13/12
   

Example-8 :-  Find the value of sin 31π/3.

Solution :-
  We know that values of sin x repeats after an interval of 2π. 
  Therefore,
  sin 31π/3 = sin(10π + π/3)
            = sin π/3
            = √3/2  
   

Example-9 :-  Find the value of cos (–1710°).

Solution :-
  We know that values of cos x repeats after an interval of 2π or 360°. 
  Therefore, 
  cos (–1710°) = cos (–1710° + 5 x 360°) 
               = cos (–1710° + 1800°) 
               = cos 90° 
               = 0
    

Example-10 :-  Prove that: 3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4 = 1

Solution :-
  L.H.S.
  3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4
= 3 x 1/2 x 2 - 4sin(π - π/6) x 1
= 3 - 4sin π/6
= 3 - 4 x 1/2
= 3 - 2
= 1 
= R.H.S
   

Example-11 :-  Find the value of sin 15°.

Solution :-
  sin 15° = sin (45° – 30°) 
          = sin 45° cos 30° – cos 45° sin 30°
          = 1/√2 x √3/2 - 1/√2 x 1/2
          = √3/2√2 - 1/2√2
          = (√3 - 1)/√2
   

Example-12 :-  Find the value of tan 13π/12.

Solution :-
  tan 13π/12 = tan(π + π/12)
             = tan π/12
             = tan(π/4 - π/6)
             = (tan π/4 - tan π/6)/(1 + tan π/4 . tan π/6)
             = (1 - 1/√3)/(1 + 1/√3)
             = (√3 - 1)/(√3 + 1)
  By rationalising,
             = (√3 - 1)/(√3 + 1) x (√3 - 1)/(√3 - 1)
             = (√3 - 1)²/[(√3)² - (1)²]
             = 2 - √3
    

Example-13 :-  Prove that: trigonometry

Solution :-
trigonometry
   

Example-14 :-  Show that: tan 3x . tan 2x . tan x = tan 3x – tan 2x – tan x

Solution :-
  We know that 3x = 2x + x.
  Therefore, tan 3x = tan (2x + x)
  tan 3x =  (tan 2x + tan x)/(1 - tan 2x . tan x)
  tan 3x – tan 3x tan 2x tan x = tan 2x + tan x  
  tan 3x – tan 2x – tan x = tan 3x tan 2x tan x  
  tan 3x tan 2x tan x = tan 3x – tan 2x – tan x
   

Example-15 :-  Prove that: cos(π/4 + x) + cos(π/4 - x) = √2 cos x

Solution :-
trigonometry
    

Example-16 :-  Prove that: trigonometry

Solution :-
trigonometry
   

Example-17 :-  Prove that: trigonometry

Solution :-
trigonometry
   

Example-18 :-  Find the principal solutions of the equation sin x = √3/2

Solution :-
  We know that sin π/3 = √3/2 and sin 2π/3 = (π - π/3) = √3/2
  Therefore, principal solutions are π/3 and 2π/3.
    

Example-19 :-  Find the principal solutions of the equation tan x = -1/√3

Solution :-
  We know that tan π/6 = 1/√3 
  tan(π - π/6) = -tan π/6 = -1/√3
  and tan(2π - π/6) = -tan π/6 = -1/√3
  Thus, tan 5π/6 = tan 11π/6 = -1/√3
  Therefore, principal solutions are 5π/6 and 11π/6.
   

Example-20 :-  Find the solution of sin x = -√3/2

Solution :-
  Given:
  sin x = -√3/2
  sin x = -sin π/3
  sin x = sin(π + π/3)
  sin x = sin 4π/3
  which gives,
  x = nπ + (-1)ⁿ 4π/6, where  n ∈ Z.
   

Example-21 :-  Solve cos x = 1/2

Solution :-
  Given:
  cos x = 1/2
  cos x = cos π/3
  Therefore, 
  x = 2nπ ± π/3, where  n ∈ Z.
    

Example-22 :-  Solve tan 2x = -cot(x + π/3).

Solution :-
  Given:
  tan 2x = -cot(x + π/3)
  tan 2x = tan(π/2 + x + π/3)
  tan 2x = tan(x + 5π/6)
  Therefore,
  2x = nπ + x + 5π/6 where  n ∈ Z. or 
   x = nπ + 5π/6 where  n ∈ Z. 
   

Example-23 :-  Solve sin 2x – sin 4x + sin 6x = 0.

Solution :-
  Given:
  sin 2x – sin 4x + sin 6x = 0
  (sin 6x + sin 2x) – sin 4x = 0
  2sin 4x cos 2x - sin 4x = 0
  sin 4x (2cos 2x - 1) = 0
  Therefore,
  sin 4x = 0 or 2cos 2x = 1
  sin 4x = 0 or  cos 2x = 1/2
  4x = nπ or 2x = 2nπ ± π/3, where  n ∈ Z.
   x = nπ/4 or x = nπ ± π/6, where  n ∈ Z.
   

Example-24 :-  Solve 2 cos² x + 3 sin x = 0

Solution :-
  Given:
  2 cos² x + 3 sin x = 0
  2 (1 - sin²x) + 3 sin x = 0
  2 - 2 sin²x + 3 sin x = 0
  2 sin²x - 3 sin x - 2 = 0
  (2sin x + 1)(sin x - 2) = 0
  Hence, sin x = -1/2 or sin x = 2
  But, sin x = 2 is not possible.  
  Therefore, 
  sin x = -1/2 = sin 7π/6
  x = nπ + (-1)ⁿ 7π/6, where  n ∈ Z.
    
CLASSES

Connect with us:

Copyright © 2015-17 by a1classes. All Rights Reserved.

www.000webhost.com