Problem 1 :
If x cos θ = y cos (θ + 2π/3) = z cos (θ + 4π/3), find the value of xy + yz + zx.
Solution :
Given that :
x cos θ = y cos (θ + 2π/3) = z cos (θ + 4π/3)
Note that xy + yz + zx = xyz(1/x + 1/y +1/z)
If we put x cos θ = y cos (θ+2π/3) = zcos(θ+4π/3) = k
Then
x cos θ = k x=k/cos θ |
ycos(θ + 2π/3) = k y = k/cos(θ+ 2π/3) |
zcos(θ+4π/3) = k z =k/cos(θ+4π/3) |
1/x = cos θ /k ---(1)
1/y = cos(θ+ 2π/3)/k ---(2)
1/z = cos(θ+4π/3)/k ---(3)
(1) + (2) + (3)
(1/x) + (1/y) + (1/z) =
(1/k) [ cos θ + cos(θ+ 2π/3) + cos(θ+ 4π/3) ]
= (1/k) [ cos θ + cos θ cos 2π/3 - sin θ sin 2π/3 + cos θ cos 4π/3 - sin θ sin 4π/3]
cos (2π/3) = -1/2 sin (2π/3) = √3/2 |
cos (4π/3) = - 1/2 sin (4π/3) = -√3/2 |
= 0
So, the value of xy + yz + zx is 0.
Problem 2 :
Prove that :
sin(A + B) sin(A − B) = sin2 A − sin2 B
Solution :
sin (A + B) = sin A cos B + cos A sin B
sin (A - B) = sin A cos B - cos A sin B
Then, sin (A + B) sin (A - B) :
= (sin A cos B + cos A sin B)(sin A cos B - cos A sin B)
= sin2A cos2B - sin A sin B cos A cos B + sin A sin B cos A cos B - cos2A sin2B
= sin2A cos2B - cos2A sin2B
= sin2A (1-sin2B) - (1-sin2A) sin2B
= sin2A - sin2A sin2B - sin2B + sin2A sin2B
= sin2A - sin2B
Problem 3 :
Prove that :
cos(A + B) cos(A − B) = cos2 A − sin2 B = cos2 B − sin2 A
Solution :
cos(A + B) = cos A cos B - sin A sin B
cos(A - B) = cos A cos B + sin A sin B
Then, cos(A + B) cos(A - B) :
= (cos A cos B - sin A sin B)(cos A cos B + sin A sin B)
= (cos2 A cos2 B + cos A sin A cos B sin B - cos A sin A cos B sin B - sin2 A sin2 B
= cos2 Acos2 B-sin2 A sin2 B -----(1)
From (1),
= cos2 A(1-sin2 B)-(1 - cos2 A)sin2 B
= cos2 A-cos2 Asin2 B - sin2 B + cos2 Asin2 B
= cos2 A - sin2 B
From (1),
= (1-sin2A) cos2 B-sin2 A (1-cos2 B)
= cos2 B-sin2A cos2 B - sin2 A + sin2A cos2 B
= cos2 B - sin2 A
Hence proved.
Problem 4 :
Prove that :
sin2(A + B) - sin2(A - B) = sin2Asin 2B
Solution :
From (i) we have
sin2 A − sin2 B = sin(A + B) sin(A − B) ----(1)
By comparing the given question with the formula, we get A = A + B and B = A - B
A + B ==> A + B + A - B ==> 2A
A - B ==> A + B - (A - B) ==> 2B
Substitute 2A for (A + B) and 2B for (A - B) in (1).
(1)-----> sin2(A + B) − sin2(A − B) = sin2Asin 2B
Hence proved.
Problem 5 :
Prove that :
cos 8θ cos 2θ = cos2 5θ − sin2 3θ
Solution :
We may write 8θ as (5θ + 3θ) and 2θ as (5θ - 3θ).
Then,
cos 8θ cos 2θ = cos (5θ + 3θ) cos (5θ - 3θ)
From (ii), we have
cos(A + B) cos(A − B) = cos2 A − sin2 B
Then,
cos (5θ + 3θ) cos (5θ - 3θ) = cos2 5θ − sin2 3θ
Hence proved.
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