Question 1. Express each of the following as the sum or difference of sines and cosines:
(i) 2 sin 3θ cos θ
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
(ii) 2 cos 3θ 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
(iii) 2 sin 4θ sin 3θ
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
(iv) 2 cos 7θ cos 3θ
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
Question 2. Prove that:
(i)
Solution:
By using the trigonometric identity,
2 sin A sin B = cos (A-B) – cos (A+B)
Taking A =
and B =
Hence, LHS = RHS
(ii)
Solution:
By using the trigonometric identity,
2 cos A cos B = cos (A+B) + cos (A-B)
Taking A =
and B =
Hence, LHS = RHS
(iii)
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
Question 3. Show that:
(i) sin 50° cos 85° =
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
(ii) sin 25° cos 115° =
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
Question 4. Prove that :
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
Question 5. Prove that :
(i) cos 10° cos 30° cos 50° cos 70° =
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
(ii) cos 40° cos 80° cos 160° =
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
(iii) sin 20° sin 40° sin 80° =
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
(iv) cos 20° cos 40° cos 80° =
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
(v) tan 20° tan 40° tan 60° tan 80° = 3
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
(vi) tan 20° tan 30° tan 40° tan 80° = 1
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
(vii) sin 10° sin 50° sin 60° sin 70° =
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
(viii) sin 20° sin 40° sin 60° sin 80° =
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
Question 6. Show that
(i) sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = 0
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
(ii) sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = 0
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
Question 7. Prove that : tan θ tan (60°-θ) tan (60°+θ) = tan 3θ
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
Question 8. If α + β = 90°, show that the maximum value of cos(α) cos(β) is
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