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:
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,![]()
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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![]()
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