Class 11 - Mathematics
Trigonometric Function - Exercise 3.3
Top Block 1
Question 1:
sin2 π/6 + cos2 π/3 – tan2 π/4 = -1/2
Answer:
LHS:
sin2 π/6 + cos2 π/3 – tan2 π/4
= (1/2)2 + (1/2)2 – 12
= 1/4 + 1/4 – 1
= 1/2 – 1
= -1/2
= RHS
Question 2:
2sin2 π/6 + cosec2 7π/6 * cos2 π/3 = 3/2
Answer:
LHS:
2sin2 π/6 + cosec2 7π/6 * cos2 π/3
= 2(1/2)2 + cosec2 (π + π/6) * (1/2)2
= 2/4 + (-cosec2 π/6) * 1/4
= 1/2 + (-2)2 * 1/4
= 1/2 + 4 * 4
= 1/2 + 1
= 3/2
= RHS
Question 3:
Prove that: cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6
Answer:
LHS:
cot2 π/6 + cosec 5π/6 + 3 tan2 π/6
= (√3)2 + cosec (π – π/6) + 3(1/√3)2
= 3 + cosec π/6 + 3 * 1/3
= 3 + 2 + 1
= 6
= RHS
Question 4:
Prove that: 2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3 = 6
Answer:
LHS:
2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3
= 2[sin (π – π/4)]2 + 2(1/√2)2 + 2 * 22
= 2[sin π/4]2 + 2 * 1/2 + 2 * 4
= 2(1/√2)2 + 1 + 8
= 2 * 1/2 + 1 + 8
= 1 + 1 + 8
= 10
= RHS
Question 5:
Find the value of:
(i) sin 75° (ii) tan 15°
Answer:
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30° [sin (x + y) = sin x cos y + cos x sin y]
= (1/√2) * (√3/2) + (1/√2) * (1/2)
= √3/2√2 + 1/2√3
= (√3 + 1)/2√3
(ii) tan 15° = tan(45° – 30°)
= (tan 45° – tan 30°)/(1 + tan 45° * tan 30°)
= (1 – 1/√3)/(1 + 1 * 1/√3)
= (1 – 1/√3)/(1 + 1/√3)
= {(√3 – 1)/√3}/{(√3 + 1)/√3}
= (√3 – 1)/(√3 + 1)
= {(√3 – 1)/(√3 + 1)} * {(√3 – 1)/ (√3 – 1)}
= (√3 – 1)2/{(√3)2 – 1)}
= (3 + 1 – 2√3)/(3 – 1)
= (4 – 2√3)/2
= 2 – √3
Question 6:
Prove that: cos(π/4 – x) * cos(π/4 – y) – sin(π/4 – x) * sin(π/4 – y) = sin(x + y)
Answer:
LHS:
cos(π/4 – x) * cos(π/4 – y) – sin(π/4 – x) * sin(π/4 – y)
= (1/2)[2 cos(π/4 – x) * cos(π/4 – y)] + (1/2)[2 sin(π/4 – x) * sin(π/4 – y)]
= (1/2)[cos{(π/4 – x) + (π/4 – y)} + cos{(π/4 – x) – (π/4 – y)}]
+ (1/2)[cos{(π/4 – x) + (π/4 – y)}] – cos{(π/4 – x) – (π/4 – y)}]
= 2 * (1/2)[cos{(π/4 – x) + (π/4 – y)}
= cos[π/2 – (x + y)]
= sin(x + y)
= RHS
Question 7:
Prove that: tan(π/4 + x)/ tan(π/4 – x) = {(1 + tan x)/(1 + tan x)}2
Answer:
LHS:
tan(π/4 + x)/tan(π/4 – x) = {(tan π/4 + tan x)/(1 – tan π/4 * tan x)}
{(tan π/4 – tan x)/(1 + tan π/4 * tan x)}
= {(1 + tan x)/(1 – tan x)}/ {(1 – tan x)/(1 + tan x)}
= {(1 + tan x)/(1 – tan x)}2
= RHS
Mddle block 1
Question 8:
Prove that: {cos(π + x) cos(-x)}/{sin(π – x) cos(π/2 + x)} = cot2 x
Answer:
LHS:
{cos(π + x) cos(-x)}/{sin(π – x) cos(π/2 + x)}
= {- cos x * cos x}/{sin x * (-sin x)}
= (-cos2 x)/(-sin2 x)
= cot2 x
= RHS
Question 9:
cos(3π/2 + x) cos(2π + x)[cot(3π/2 – x) + cot(2π + x)] = 1
Answer:
LHS:
cos(3π/2 + x) cos(2π + x)[cot(3π/2 – x) + cot(2π + x)]
= sin x * cos x[tan x + cot x]
= sin x * cos x[sin x/cos x + cos x/sin x]
= sin x * cos x[(sin2 x + cos2 x)/(cos x * sin x)]
= 1
= RHS
Question 10:
Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Answer:
L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
= (1/2)[2sin(n + 1)x sin (n + 2)x + 2cos (n + 1)x cos (n + 2)x]
= (1/2)[cos{(n + 1)x – (n + 2)x} – cos{(n + 1)x + (n + 2)x} + [cos{(n + 1)x + (n + 2)x}] + cos{(n + 1)x –
(n + 2)x}]
= (1/2) * 2 cos{(n + 1)x – (n + 2)x}
= cos(-x)
= cos x
= RHS
Question 11:
Prove that: cos(3π/4 + x) – cos(3π/4 – x) = -√2sin x
Answer:
LHS:
cos(3π/4 + x) – cos(3π/4 – x)
= -2sin[{(3π/4 + x) + (3π/4 – x)}/2] * sin[{(3π/4 + x) – (3π/4 – x)}/2]
= -2sin 3π/4 * sin x
= -2sin (π – π/4) * sin x
= -2sin π/4 * sin x
= -2 * 1/√2 * sin x
= -√2 sin x
= RHS
Question 12:
Prove that sin2 6x – sin2 4x = sin 2x sin 10x
Answer:
L.H.S.
sin2 6x – sin2 4x
= (sin 6x + sin 4x) (sin 6x – sin 4x)
= [2 sin(6x + 4x)/2 * cos (6x – 4x)/2][2 cos(6x + 4x)/2 * sin (6x – 4x)/2]
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2sin x cos x)
= sin 10x sin 2x
= R.H.S.
Question 13:
Prove that cos2 2x – cos2 6x = sin 4x sin 8x
Answer:
L.H.S.
cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2 cos(2x + 6x)/2 * cos (2x – 6x)/2][-2 sin(2x + 6x)/2 * sin (2x – 6x)/2]
= [2 cos 4x cos 2x] [–2 sin 4x sin(-2x)]
= [2 sin 4x cos (-4x)][-2 sin 4x sin(-2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
Question 14:
Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Answer 14:
L.H.S.
sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= 2 sin(2x + 6x)/2 * cos (2x – 6x)/2 + 2sin 4x
= 2 sin 4x cos (– 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1 + 1)
= 2 sin 4x (2 cos2 x)
= 4 cos2 x sin4x
= R.H.S.
Question 15:
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer:
L.H.S:
cot 4x (sin 5x + sin 3x)
= (cos 4x/sin 4x)[2sin (5x + 3x)/2 * cos (5x – 3x)/2]
= (cos 4x/sin 4x)[2sin 4x * cos x]
= 2 cos 4x cos x
RHS:
cot x (sin 5x – sin 3x)
= (cos x/sin x)[2cos (5x + 3x)/2 * sin (5x – 3x)/2]
= (cos x/sin x)[2cos 4x * sin x]
= 2 cos 4x cos x
So, L.H.S. = R.H.S.
Question 16:
Prove that: (cos 9x – cos 5x)/(sin 17x – sin 3x) = -sin 2x/cos 10x
Answer:
LHS:
(cos 9x – cos 5x)/(sin 17x – sin 3x)
= [-2 sin(9x + 5x)/2 * sin(9x – 5x)/2]/ [2 cos(17x + 3x)/2 * sin(17x – 3x)/2]
= (-2 sin 7x * sin 2x)/(2 cos 10x * sin 7x)
= -sin 2x/cos 10x
= RHS
Question 17:
Prove that: (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x
Answer:
LHS:
(sin 5x + sin 3x)/(cos 5x + cos 3x)
= {2 * sin (5x + 3x)/2 * cos (5x – 3x)/2}/{2 * cos (5x + 3x)/2 * cos (5x – 3x)/2}
= {2*sin 4x * cos x}/{2*cos 4x * cos x}
= sin 4x/ cos 4x
= tan 4x = RHS
Question 18:
Prove that: (sin x – sin y)/(cos x + cos y) = tan (x – y)/2
Answer:
LHS:
(sin x – sin y)/(cos x + cos y)
= {2 * cos (x + y)/2 * sin (x – y)/2}/{2 * cos (x + y)/2 * cos (x – y)/2}
= {sin (x – y)/2}/{cos (x – y)/2}
= tan (x – y)/2
= RHS
Question 19:
Prove that: (sin x + sin 3x)/(cos x + cos 3x) = tan 2x
Answer:
LHS:
(sin x + sin 3x)/(cos x + cos 3x)
= {2 * sin (x + 3x)/2 * cos (x – 3x)/2}/{2 * cos (x + 3x)/2 * cos (x – 3x)/2}
= sin 2x/cos 2x
= tan 2x
= RHS
Question 20:
Prove that: (sin x – sin 3x)/(sin2 x – cos2 x) = 2 sin x
Answer:
LHS:
(sin x – sin 3x)/(sin2 x – cos2 x)
= {2 * cos (x + 3x)/2 * sin (x – 3x)/2}/(-cos 2x)
= {2 * cos 2x * sin(-x)}/ (-cos 2x)
= -2 * sin(-x)
= 2 sin x
= RHS
Question 21:
Prove that: (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x
Answer:
LHS:
(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x)
= {(cos 4x + cos 2x) + cos 3x}/{(sin 4x + sin 2x) + sin 3x}
= {2 * cos (4x + 2x)/2 * cos (4x – 2x)/2 + cos 3x}/{2 * sin (4x + 2x)/2 * cos (4x – 2x)/2 + sin 3x}
= {2 * cos 3x * cos x + cos 3x}/{2 * sin 3x * cos x + sin 3x}
= {cos 3x (2 cos x + 1)}/{sin 3x (2 cos x + 1)}
= cot 3x
= RHS
Question 22:
Prove that: cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Answer:
LHS:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x + x) (cot 2x + cot x)
= cot x cot 2x – {(cot 2x * cot x – 1)/(cot x + cot 2x)}(cot 2x + cot x)
= cot x cot 2x – (cot 2x cot x – 1)
= cot x cot 2x – cot 2x cot x + 1
= 1
= R.H.S.
Question 23:
Prove that: tan 4x = {4 tan x (1 – tan2 x)}/( 1 – 6tan2 x + tan4 x)
Answer:
LHS:
tan 4x
= tan 2(2x)
= 2 tan 2x/(1 – tan2 2x)
= [2{2tan x/(1 – tan2 x)}]/[1 – {2tan x/(1 – tan2 x)}2]
= [4tan x/(1 – tan2 x)]/[1 – 4tan2 x/(1 – tan2 x)2]
= [4tan x/(1 – tan2 x)]/[{(1 – tan2 x)2 – 4tan2 x}/(1 – tan2 x)2]
= [4tan x(1 – tan2 x)]/[{(1 + tan4 x – 2tan2 x – 4tan2 x}]
= {4 tan x(1 – tan2 x)}/( 1 – 6tan2 x + tan4 x)
= RHS
Question 24:
Prove that: cos 4x = 1 – 8sin2 x cos2 x
Answer:
LHS: cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [Since cos 2A = 1 – 2 sin2 A]
= 1 – 2(2 sin x cos x)2 [Since sin 2A = 2sin A cos A]
= 1 – 8 sin2 x cos2 x
= RHS
Question 25:
Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Answer:
LHS:
cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [Since cos 3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 x – 1)3 – 3(2 cos2 x – 1) [Since cos 2x = 2 cos2 x – 1]
= 4 [(2 cos2 x)3 – 13 – 3(2 cos2 x)2 + 3(2 cos2 x)] – 6cos2 x + 3
= 4 [8 cos6 x – 1 – 12 cos4 x + 6 cos2 x] – 6 cos2 x + 3
= 32 cos6 x – 4 – 48 cos4 x + 24 cos2 x – 6 cos2 x + 3
= 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
= R.H.S.
Bottom Block 3
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