If cospΘ + cosqΘ = o. prove that the different values of Θ form two arithmetical progressions in which the common differences are 2π/p+q and 2π/p-q respectively.
cos px = -cos qx
cos px = cos ( pi - qx )
px = n ( pi ) + / - qx
taking the first case
px = n ( pi ) + qx
x ( p - q ) = n ( pi )
x = n ( ? p i) / ( p - q )
taking the second case we get
x = n ( ? pi ) / ( p + q )
We find the common difference is what is asked.
To prove that the different values of Θ form two arithmetic progressions, let's first manipulate the given equation:
cospΘ + cosqΘ = 0
Rearranging this equation, we get:
cospΘ = -cosqΘ
Now, recalling the trigonometric identity:
cos(θ + π) = -cosθ
We can observe that if we set:
Θ = (2n + 1)π - qΘ
where n is an integer, the equation will hold true.
Next, we substitute Θ with the given value:
(2n + 1)π - qΘ = pΘ
Expanding this equation, we have:
(2n + 1)π = (q + p)Θ
Dividing both sides by (q + p), we find:
Θ = (2n + 1)π / (q + p)
This equation provides the general solution for Θ.
Now, let's calculate the difference between consecutive terms in each of the two arithmetic progressions.
For the first progression:
Let Θ₁ and Θ₂ be two consecutive terms. Their general expressions can be written as:
Θ₁ = (2n + 1)π / (q + p)
Θ₂ = (2n + 3)π / (q + p)
The difference between these terms is:
Δ₁ = Θ₂ - Θ₁ = [(2n + 3)π / (q + p)] - [(2n + 1)π / (q + p)]
Simplifying this expression, we get:
Δ₁ = 2π / (q + p)
Thus, the common difference for the first arithmetic progression is 2π / (q + p).
For the second progression:
Let Θ₃ and Θ₄ be two consecutive terms. Their general expressions can be written as:
Θ₃ = (2n + 1)π / (q + p)
Θ₄ = (2n - 1)π / (q + p)
The difference between these terms is:
Δ₂ = Θ₄ - Θ₃ = [(2n - 1)π / (q + p)] - [(2n + 1)π / (q + p)]
Simplifying this expression, we get:
Δ₂ = -2π / (q - p)
Thus, the common difference for the second arithmetic progression is -2π / (q - p).
To summarize, we have shown that the different values of Θ form two arithmetic progressions with common differences of 2π / (q + p) and -2π / (q - p) respectively.
To prove that the different values of Θ form two arithmetic progressions with the common differences of 2π/(p+q) and 2π/(p-q), let's start by manipulating the given equation.
Given: cos(pΘ) + cos(qΘ) = 0
Using the trigonometric identity, cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2), we can rewrite the equation as:
2cos((pΘ+qΘ)/2)cos((pΘ-qΘ)/2) = 0
Now, we have two possibilities:
1. The first factor is equal to 0:
cos((pΘ+qΘ)/2) = 0
This implies that (pΘ+qΘ)/2 = (2n+1)π/2, where n is an integer.
Simplifying, we get:
pΘ+qΘ = (2n+1)π
Θ = (2n+1)π/(p+q)
The difference between consecutive values of Θ in this case is:
Θ2 - Θ1 = [(2n+3)π/(p+q)] - [(2n+1)π/(p+q)] = 2π/(p+q)
2. The second factor is equal to 0:
cos((pΘ-qΘ)/2) = 0
This implies that (pΘ-qΘ)/2 = (2m+1)π/2, where m is an integer.
Simplifying, we get:
pΘ-qΘ = (2m+1)π
Θ = (2m+1)π/(p-q)
The difference between consecutive values of Θ in this case is:
Θ2 - Θ1 = [(2m+3)π/(p-q)] - [(2m+1)π/(p-q)] = 2π/(p-q)
Therefore, the different values of Θ form two arithmetic progressions with common differences of 2π/(p+q) and 2π/(p-q) respectively.