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VOOZH | about |
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VOOZH | about |
It is vital for the students working on Class 11 Mathematics to get a hold of the Transformation Formulae. Therefore, the formulae featured in Chapter 8 of RD Sharma’s book serve as a prerequisite for solving different trigonometric problems. This guide will lead you to Exercise 8. 1, with clear specific recommendations and reasoning.
Transforming Formulae in trigonometry are generally expressed so as to change one expression to another. Both of these formulae are quite useful in solving different equations as well as in confirming different identity equations.
Exercise 8. 1 is intended for transforming a certain number of practical tasks into the use of the transformation formulae by students. This exercise will help to clear your understanding as well as check how successfully you can manipulate with these formulae.
Solution:
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
Taking A = 3θ and B = θ
2 sin 3θ cos θ = sin (3θ+θ) + sin (3θ-θ)
= sin 4θ + sin 2θ
Solution:
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
Taking A = 2θ and B = 3θ
2 cos 3θ sin 2θ = sin (3θ+2θ) + sin (2θ-3θ)
= sin 5θ + sin (-θ)
= sin 5θ - sin θ
Solution:
By using the trigonometric identity,
2 sin A sin B = cos (A-B) - cos (A+B)
Taking A = 4θ and B = 3θ
2 sin 4θ sin 3θ = cos (4θ-3θ) - cos (4θ+3θ)
= cos θ - cos 7θ
Solution:
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
Taking A = 7θ and B = 3θ
2 cos 7θ cos 3θ = cos (7θ+3θ) + cos (7θ-3θ)
= cos 10θ - cos 4θ
Solution:
By using the trigonometric identity,
2 sin A sin B = cos (A-B) - cos (A+B)
Taking A = and B =
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
Taking A = and B =
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
Taking A = and B =
= sin + sin
= 1 +
=
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = (sin (A+B) + sin (A-B))
Taking A = 50° and B = 85°
sin 50° cos 85° = (sin (50°+85°) + sin (50°-85°))
= (sin (135°) + sin (-35°))
= (sin (180°-45°) - sin (35°)) (sin(-θ)=-sin θ)
= (sin (45°) - sin (35°)) (sin(π-θ)=sin θ)
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = (sin (A+B) + sin (A-B))
Taking A = 25° and B = 115°
sin 25° cos 115° = (sin (25°+85°) + sin (25°-115°))
= (sin (140°) + sin (-90°))
= (sin (180-40°) - sin (90°)) (sin(-θ)=-sin θ)
= (sin (40°) - sin (90°)) (sin(π-θ)=sin θ)
= (sin (40°) - 1)
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
Taking A = +θ and B = -θ
Using the identity again, we have
Taking A = 2θ and B = θ
Hence, LHS = RHS
Solution:
cos 10° cos 30° cos 50° cos 70° = cos 30° cos 10° cos 50° cos 70°
= (cos 10° cos 50°) cos 70°
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
Taking A = 10° and B = 50°
= ([cos (10°+50°) + cos (10°-50°)]) cos 70°
= (cos (60°) + cos (-40°)) cos 70°
= ( + cos (40°)) cos 70°
= cos 70° + (cos 70° cos (40°))
Again using the identity, we get
= cos 70° + ([cos (70°+40°) + cos (70°-40°)])
= cos 70° + [cos (110°) + cos (30°)]
= cos 70° + [cos (110°) + ]
= cos 70° + cos (110°) +
= (cos 70° + cos (110°)) +
= (cos 70° + cos (180°-70°)) +
= (cos 70° - cos (70°)) +
=
Hence, LHS = RHS
Solution:
cos 40° cos 80° cos 160° = cos 80° (cos 40° cos 160°)
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
Taking A = 160° and B = 40°
= cos 80° ([cos (160°+40°) + cos (160°-40°)])
= cos 80° ([cos (200°) + cos (120°)])
= cos 80° ([cos (180°+20°) + cos (180°-60°)])
= cos 80° ([- cos (20°) + (-cos (60°))])
= cos 80° ([- cos (20°) - cos (60°)])
= cos 80° ([- cos (20°) - ])
= (cos 80° cos (20°) + cos 80°])
Again using the identity, we get
Hence, LHS = RHS
Solution:
sin 20° sin 40° sin 80° = (sin 20° sin 40°) sin 80°
By using the trigonometric identity,
2 sin A sin B = cos (A-B) - cos (A+B)
sin A sin B = [cos (A-B) - cos (A+B)]
Taking A = 40° and B = 20°
= ([cos (40°-20°) - cos (40°+20°)]) sin 80°
= sin 80° [cos (20°) - cos (60°)]
= sin 80° [cos (20°) - ]
= [sin 80° cos (20°) - sin 80°]
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = [sin (A+B) + sin (A-B)]
Taking A = 80° and B = 20°
Hence, LHS = RHS
Solution:
cos 20° cos 40° cos 80° = cos 40° (cos 20° cos 80°)
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
Taking A = 80° and B = 20°
Again using the identity, we get
Hence, LHS = RHS
Solution:
tan 20° tan 40° tan 60° tan 80° = tan 60°
=
=
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
and, 2 sin A sin B = cos (A-B) - cos (A+B)
sin A sin B = [cos (A-B) - cos (A+B)]
Taking A = 40° and B = 20°
Again using the identity, we get
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = [sin (A+B) + sin (A-B)]
Taking A = 80° and B = 20°
Hence, LHS = RHS
Solution:
tan 20° tan 30° tan 40° tan 80° = tan 30°
=
=
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
and, 2 sin A sin B = cos (A-B) - cos (A+B)
sin A sin B = [cos (A-B) - cos (A+B)]
Taking A = 40° and B = 20°
Again using the identity, we get
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = \frac{1}{2}[sin (A+B) + sin (A-B)]
Taking A = 80° and B = 20°
Hence, LHS = RHS
Solution:
sin 10° sin 50° sin 60° sin 70° = sin 60° (sin 10° sin 50° sin 70°)
= \frac{\sqrt{3}}{2} (sin (90-80°) sin (90-40°) sin (90-20°))
= (cos (80°) cos (40°) cos (20°))
= cos 40° (cos 80° cos 20°)
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
Taking A = 80° and B = 20°
Again using the identity, we get
Hence, LHS = RHS
Solution:
sin 20° sin 40° sin 60° sin 80° = sin 60° (sin 20° sin 40° sin 80°)
= (sin 20° sin 40°) sin 80°
By using the trigonometric identity,
2 sin A sin B = cos (A-B) - cos (A+B)
sin A sin B = [cos (A-B) - cos (A+B)]
Taking A = 40° and B = 20°
By using the trigonometric identity,
2 sin A cos B = sin (A+B) + sin (A-B)
sin A cos B = [sin (A+B) + sin (A-B)]
Taking A = 80° and B = 20°
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 sin θ sin Φ = cos (θ-Φ) - cos (θ+Φ)
sin θ sin Φ =[cos (θ-Φ) - cos (θ+Φ)]
sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = ([cos (A-(B-C)) - cos (A+(B-C))]) + ([cos (B-(C-A)) - cos (B+(C-A))]) + ([cos (C-(A-B)) - cos (C+(A-B))])
= (cos (A-B+C)) - cos (A+B-C) + cos (B-C+A) - cos (B+C-A) + cos (C-A+B) - cos (C+A-B))
= (cos (A-B+C)) - cos (C+A-B) - cos (A+B-C) + cos (B-C+A) - cos (B+C-A) + cos (C-A+B))
=
= 0
Hence, LHS = RHS
Solution:
By using the trigonometric identity,
2 sin θ cos Φ = sin (θ+Φ) + sin (θ-Φ)
sin θ cos Φ = [sin (θ+Φ) + sin (θ-Φ)]
sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = ([sin (B-C+(A-D)) + sin (B-C-(A-D))]) + ([sin (C-A+(B-D)) + sin (C-A-(B-D))]) +([sin (A-B+(C-D)) + sin (A-B-(C-D))])
= ([sin (A+B-C-D) + sin (-A+B-C+D)]) + ([sin (-A+B+C-D) + sin (-A-B+C+D)]) +([sin (A-B+C-D) + sin (A-B-C+-D)])
= (sin (A+B-C-D) + sin (-(A-B+C-D)) + sin (-(A-B-C+D)) + sin (-(A+B-C-D)) +sin (A-B+C-D) + sin (A-B-C+-D))
=(sin (A+B-C-D) - sin(A-B+C-D) - sin (A-B-C+D) - sin (A+B-C-D) +sin (A-B+C-D) + sin (A-B-C+-D))
=
= 0
Hence, LHS = RHS
Solution:
tan θ tan (60°-θ) tan (60°+θ) = tan θ (tan (60°-θ)) (tan (60°+θ))
By using the trigonometric identity,
tan (a+b) =
tan (a+b) =
= tan 3θ
Hence, LHS = RHS
Solution:
cos(α) cos(β) = y
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
cos A cos B = [cos (A+B) + cos (A-B)]
Taking A = α and B = β
cos(α) cos(β) = [cos (α+β) + cos (α-β)]
As, α + β = 90°
y = [cos (90°) + cos (α-β)]
y = [0 + cos (α-β)]
y = (cos (α-β))
AS, we know that range of cos function is [-1,1]
Hence, the maximum value of cos(α) cos(β) is
