Let ƒÆ=sin −1 7/25 . Consider the sequence of values defined by a n =sin(nƒÆ) . They satisfy the recurrence relation
a n+2 =k 1 a n+1 +k 0 a n ,n�¸N
for some (fixed) real numbers k 1 ,k 0 . The sum k 1 +k 0 can be written as p q , where p and q are positive coprime integers. What is the value of p+q ?
The first step to find the value of p and q is to simplify the given recurrence relation aₙ₊₂ = k₁aₙ₊₁ + k₀aₙ.
To simplify the notation, let's replace the angle symbol "ƒÆ" with "θ" to represent sin⁻¹(7/25).
We know that sin(θ) = 7/25. Therefore, θ = sin⁻¹(7/25).
Now, let's substitute aₙ = sin(nθ) into the recurrence relation:
sin((n+2)θ) = k₁sin((n+1)θ) + k₀sin(nθ)
Next, we can use the trigonometric identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B) to expand the right side of the equation:
sin((n+2)θ) = k₁sin(nθ)cos(θ) + k₁cos(nθ)sin(θ) + k₀sin(nθ)
Using the identity sin(θ) = 7/25, we can simplify the equation further:
sin((n+2)θ) = (k₁(7/25)cos(θ) + k₀(7/25))(sin(nθ))
Comparing the equation above with the form aₙ₊₂ = Aaₙ₊₁ + Baₙ, we can deduce that:
A = k₁(7/25)cos(θ) + k₀(7/25)
B = 1
Now, let's find the values of A and B:
A = k₁(7/25)cos(θ) + k₀(7/25)
= (7k₁cos(θ) + 7k₀) / 25 * (1/1)
= (7k₁cos(θ) + 7k₀) / 25
B = 1
From the given recurrence relation aₙ₊₂ = k₁aₙ₊₁ + k₀aₙ, we know that p = A and q = B.
Therefore, p + q = (7k₁cos(θ) + 7k₀) / 25 + 1
= (7k₁cos(θ) + 7k₀ + 25) / 25
The value of p + q is (7k₁cos(θ) + 7k₀ + 25) / 25.
However, unless the values of k₁ and k₀ are specified, we cannot determine the exact value of p + q.