prove that sin(n+1)A-sin(n-1)A/cos(n+1)A+2cos(nA)+cos(n-1)A=tanA/2
To prove the given trigonometric identity, we will make use of the trigonometric identities for the sum and difference of angles, as well as the double-angle identity for tangent.
Let's start by rewriting the left side of the equation:
(sin(n+1)A - sin(n-1)A) / (cos(n+1)A + 2cos(nA) + cos(n-1)A)
Using the sum and difference of angles identities, we can rewrite the numerator and denominator as follows:
(sin(nA)cos(A) + cos(nA)sin(A) - (sin(nA)cos(A) - cos(nA)sin(A))) / (cos(nA)cos(A) - sin(nA)sin(A) + 2cos(nA) + cos(nA)cos(A) + sin(nA)sin(A))
Simplifying the numerator and denominator, we have:
(2cos(nA)sin(A)) / (2cos(nA) + 2cos(nA))
Canceling out the common terms of 2cos(nA), we have:
(sin(A)) / cos(A)
Using the identity for tangent (tan(x) = sin(x) / cos(x)), we can rewrite it as:
tan(A)
Finally, using the double-angle identity for tangent (tan(2x) = 2tan(x) / (1 - tan^2(x))), we can write:
tan(A) / 2
Therefore, we have proven that:
(sin(n+1)A - sin(n-1)A) / (cos(n+1)A + 2cos(nA) + cos(n-1)A) = tan(A) / 2