Prove that
sin(n+1)A-sin(n-1)A/cos(n+1)A+2cosnA+cos(n-1)A = tanA/2
To prove the given equation, we'll start by applying the sine addition formula for sin(n+1)A and sin(n-1)A:
sin(n+1)A = sin(nA + A)
= sin(nA)cos(A) + cos(nA)sin(A)
sin(n-1)A = sin(nA - A)
= sin(nA)cos(A) - cos(nA)sin(A)
By substituting these expressions into the given equation, we get:
(sin(n+1)A - sin(n-1)A) / (cos(n+1)A + 2cos(nA) + cos(n-1)A) = tan(A/2)
Let's simplify the left side of the equation:
[ (sin(nA)cos(A) + cos(nA)sin(A)) - (sin(nA)cos(A) - cos(nA)sin(A)) ] / (cos(n+1)A + 2cos(nA) + cos(n-1)A) = tan(A/2)
Cancelling out the terms:
2cos(nA)sin(A) / (cos(n+1)A + 2cos(nA) + cos(n-1)A) = tan(A/2)
Now, let's focus on the denominator and expand it using the cosine addition formula for cos(n+1)A and cos(n-1)A:
cos(n+1)A = cos(nA + A)
= cos(nA)cos(A) - sin(nA)sin(A)
cos(n-1)A = cos(nA - A)
= cos(nA)cos(A) + sin(nA)sin(A)
Substituting these values back into our equation, we have:
2cos(nA)sin(A) / (cos(nA)cos(A) - sin(nA)sin(A) + 2cos(nA) + cos(nA)cos(A) + sin(nA)sin(A))
By combining like terms:
2cos(nA)sin(A) / (2cos(nA) + 2cos(nA))
= 2cos(nA)sin(A) / 2(cos(nA) + cos(nA))
= cos(nA)sin(A) / (cos(nA) + cos(nA))
Using the identity sin(2θ) = 2sin(θ)cos(θ), we can further simplify:
2sin(A)cos(nA) / 2cos(nA)
= sin(A) / cos(nA)
= tan(A)
Finally, we need to prove that tan(A) is equal to tan(A/2):
Using the half-angle formula for the tangent:
tan(A/2) = sin(A) / (1 + cos(A))
We can see that tan(A) = sin(A) / cos(nA) is equivalent to tan(A/2). Hence, the given equation is proven:
(sin(n+1)A - sin(n-1)A) / (cos(n+1)A + 2cos(nA) + cos(n-1)A) = tan(A/2)