Question-1 :- Prove that: 2 cos π/13 . cos 9π/13 + cos 3π/13 + cos 5π/13 = 0

Solution :-

Question-2 :-  Prove that: (sin 3x + sin x) . sin x + (cos 3x – cos x) . cos x = 0

Solution :-

Question-3 :- Prove that: (cos x + cos y)² + (sin x – sin y)² = 4 cos² (x + y)/2

Solution :-

Question-4 :- Prove that: (cos x - cos y)² + (sin x – sin y)² = 4 sin² (x - y)/2

Solution :-

Question-5 :-  Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x . cos 2x . sin 4x

Solution :-

Question-6 :- Prove that:

Solution :-

Question-7 :- Prove that: sin 3x + sin 2x – sin x = 4sin x . cos x/2 . cos 3x/2

Solution :-

Question-8 :-  If tan x = -4/3, x in 2nd quadrant, find the value of sin x/2, cos x/2 and tan x/2.

Solution :-

  Since π/2 < x < π (2nd quadrant), cos x is negative
  Also, π/4 < x/2 < π/2 (1st quadrant)
  Therefore, sin x/2 is positive and cos x/2 is positive.
  Now,
  sec² x = 1 + tan² x 
         = 1 + (-4/3)²
         = 1 + 16/9
         = 25/9
  sec x = 5/3
  1/cos x = 5/3
  cos x = 3/5 
  Here, x lies on the second quadrant, so cos x is negative.
  cos x = -3/5
  
  Now, 2sin² x/2 = 1 - cos x
                 = 1 - (-3/5)
                 = 1 + 3/5
                 = 8/5
        sin² x/2 = 8/10
         sin x/2 = 2√2/√10 = 2/√5 (π/4 < x/2 < π/2, sin x/2 is positive)
 
  Now, cos² x/2 = 1 - sin² x/2
                = 1 - (2/√5)²
                = 1 - 4/5
       cos² x/2 = 1/5
        cos x/2 = 1/√5 (π/4 < x/2 < π/2, cos x/2 is positive)
 
  Now, tan x/2 = (sin x/2)/(cos x/2)
               = (2/√5)/(1/√5)
               = 2/√5 x (√5/1)
       tan x/2 = 2 (π/4 < x/2 < π/2, tan x/2 is positive)
    

Question-9 :-  If cos x = -1/3, x in 3rd quadrant, find the value of sin x/2, cos x/2 and tan x/2.

Solution :-

  Since π < x < 3π/2, cos x is negative
  Also, π/2 < x/2 < 3π/4
  Therefore, sin x/2 is positive and cos x/2 is negative.
  Now,
  
  
    

Question-10 :-  If sin x = 1/4, x in 2nd quadrant, find the value of sin x/2, cos x/2 and tan x/2.

Solution :-

  Since π/2 < x < π (2nd quadrant), cos x is negative
  Also, π/4 < x/2 < π/2 (1st quadrant)
  Therefore, sin x/2 is positive and cos x/2 is positive.
  Now,
  cos² x = 1 - sin² x 
         = 1 - (1/4)²
         = 1 - 1/16
         = 15/16
  cos x = √15/4
  Here, x lies on the second quadrant, so cos x is negative.
  cos x = -√15/4