Thursday, 26 June 2014

trigonometry - How prove this trigonometric identity



Show that




n1k=0tanh(x1nsin2(2k+14nπ))1+tanh2xtan2(2k+14nπ)=tanh(2nx)





Thank you ,This problem I take some hours,and at last I don't prove it



and This problem is from
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This book have some same problem.all is not true? if not true,and we how find it or edit it somewhere?
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Thank you achille hui,he told me this following maybe is true,Now How prove it?




n1k=0tanhxnsin2(2k+14nπ)1+tanh2xtan2(2k+14nπ)=tanh(2nx)



Answer



Notice




cosh(2nx)=T2n(coshx) and cos(2nx)=T2n(cosx)



where T2n(z) is a
Chebyshev polynomial of the first kind. Using the 2nd relation above, it is clear the roots of T2n(x) has the
form:
±cos(2k+14nπ), for k=0,,n1



From this, we arrive following expansion of cosh(2nx):



cosh(2nx)=An1k=0(cosh2xcos2(2k+14nπ))

for some constant A we don't care.
Taking logarithm, differentiate w.r.t x and divide by 2n for both sides, we find:
tanh(2nx)=12nn1k=02sinhxcoshxcosh2xcos2(2k+14nπ)=1nn1k=0sinhxcoshxsinh2x+sin2(2k+14nπ)=1nn1k=0tanhxtanh2x+sin2(2k+14nπ)(1tanh2x)=n1k=0tanhxnsin2(2k+14nπ)1+(1sin2(2k+14nπ)1)tanh2x=n1k=0tanhxnsin2(2k+14nπ)1+tanh2xtan2(2k+14nπ)


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