Problem with induction on trigonometry

**Apo37** · October 26th 2009, 04:22 AM

Hello, I'm stuck in solving this problem where I got to prove this equation with induction. Got problems with the 1st step already x.x On the 1st step, do I have to prove the equation for n=0 or n=1 on the following problem? How and why? On the 2nd step for n=m I know what to do... But then I'm again stuck on the 3rd step, for n=m+1, I don't know how to transform the trigonometry forms on the left side of the equation to make it the same as the right side. I'd appreciate any help or explanation. Thanks in advance. The problem is:

**red_dog** · October 26th 2009, 08:34 AM

The first step of induction is to verify the identity for the least value of n, in this case for n=1. That means

$\frac{1}{2}+\cos x=\frac{\sin\displaystyle\frac{3x}{2}}{2\sin\displ aystyle\frac{x}{2}}$

But $\frac{\sin\displaystyle\frac{3x}{2}}{2\sin\display style\frac{x}{2}}=\frac{3\sin\displaystyle\frac{x} {2}-4\sin^3\frac{x}{2}}{2\sin\displaystyle\frac{x}{2}} =$

$=\frac{3-4\sin^2\displaystyle\frac{x}{2}}{2}=\frac{1}{2}+1-2\sin^2\frac{x}{2}=\frac{1}{2}+\cos x$

Now we suppose that the identity holds for n and we have to prove for n+1. That means

$1+\sum_{k=1}^{n+1}\cos kx=\frac{\sin\displaystyle\frac{(2n+3)x}{2}}{2\sin \displaystyle\frac{x}{2}}$

We have:

$1+\sum_{k=1}^{n+1}\cos kx=1+\sum_{k=1}^n\cos kx+\cos(n+1)x=\frac{\sin\displaystyle\frac{(2n+1)x }{2}}{2\sin\displaystyle\frac{x}{2}}+\cos(n+1)x=$

$=\frac{\sin\displaystyle\frac{(2n+1)x}{2}+2\sin\fr ac{x}{2}\cos(n+1)x}{2\sin\displaystyle\frac{x}{2}} =$

$=\frac{\sin\displaystyle\frac{(2n+1)x}{2}+\sin\fra c{(2n+3)x}{2}-\sin\frac{(2n+1)x}{2}}{2\sin\displaystyle\frac{x}{ 2}}=\frac{\sin\displaystyle\frac{(2n+3)x}{2}}{2\si n\displaystyle\frac{x}{2}}$

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