Question

Prove that sin (n + 1) a -sin (n-1) a a = tan cos (n +1) a +2 cos na +cos (n-1) a 2. +
$\sin(n + 1) \alpha - \sin(n - 1) \alpha \div \cos(n + 1) \alpha + 2 \cos(n \alpha ) + \cos(n - 1) \alpha$
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Answer 1

Answer:

We check the equation for n = 2 and we take:

1/3 + 1/4 = 1 – 1/2 + 1/3 - 1/4 = 1/2 + 1/12 = 6/12 + 1/12 = 7/12, true

We assume that our equation holds for n = k and we write:

1/(k+1) + 1/(k+2) + … + 1/2k = 1 – 1/2 + 1/3 - 1/4 + … + 1/(2k-1) - 1/(2k) … (1)

We will prove that for n = k+1 we have:

1/(k+2) + 1/(k+3) + … + 1/2k + 1/(2k+1) + 1/(2k+2) = 1 – 1/2 + 1/3 - 1/4 + … + 1/(2k-1) - 1/(2k) + 1/(2k+1) - 1/(2k+2) … (2)

Now, we add [1/(2k+1) + 1/(2k+2) - 1/(k+1)] to both parts of (1) and we take:

1/(k+2) + 1/(k+3) + … + 1/2k + 1/(2k+1) + 1/(2k+2) = 1 – 1/2 + 1/3 - 1/4 + … + 1/(2k-1) - 1/(2k) + [1/(2k+1) + 1/(2k+2) - 1/(k+1)]

If we prove that 1/(2k+2) - 1/(k+1) = - 1/(2k+2) the proof of (2) will be complete.

We can easily see that:

1/(2k+2) - 1/(k+1) = [(k+1) - (2k+2)]/[(2k+2)(k+1)] = [(k+1) - 2(k+1)]/[(2k+2)(k+1)] =

[(k+1)(1–2)]/[(k+1)(2k+2)] = - 1/(2k+2) and the proof is now complete.