Prove Trigonometric Statement: Ideas & Solutions

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Have you tried mathematical induction?
 
Metaleer said:
Have you tried mathematical induction?

Hi thanks for the reply,

No, don't know why I didn't think of that,

Will give it a go now.
 
I don't think this is to be done with mathematical induction. I ended up with these massive identities and really long formulas,

Is there another way?
 
I managed to do it using

<br /> \cos a \sin b = (\sin (a+b) - \sin(a-b))/2<br />

PROBLEM SOLVED!
 
Have u proven the P1 case and related Pk with Pk+1?

Do u think the identity below will help ? :wink:

sin(x)cos[(2k+1)x]
=\frac{sin[((2k+2)x)-((2k)x)]}{2} + \frac{cos[((2k+2)x)+((2k)x)]}{2}
=\frac{1}{2}[sin((2k+2)x)-sin((2k)x))]
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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