Solve #sin((n+1)theta) = sin(ntheta)# ?

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Answer 1 · 2017-12-19T10:43:24

See below.

If #sin((n+1)theta) = sin(ntheta) rArr 1/(2i)(e^(i(n+1)theta)-e^(-i(n+1)theta)) = 1/(2i)(e^(i ntheta)-e^(-i ntheta))# then

#e^(i(n+1)theta)-e^(-i(n+1)theta) = e^(i ntheta)-e^(-i ntheta)#

Calling #y = e^(i theta)#

#y^(n+1)-y^(-(n+1)) =y^n-y^(-n)# or

#y^(n+1)-y^n = y^(-(n+1))-y^-n# or

#y^n(y-1)=y^-n(y^-1-1)# or

#y^(2n) = (y^-1-1)/(y-1)=-y^-1# then

#y^(2n+1)=-1# or equivalently

#e^(i(2n+1)theta)=-1# or

#(2n+1)theta = pi + 2k pi# and then

#theta ={( (2k+1)/(2n+1)pi) uu (theta = 0+ 2kpi)}# for #k in ZZ#

Answer 2 · 2017-12-19T11:16:33

#theta in {(2m)/npi}uu{((2m+1)/(2n+1))pi}, m in ZZ.#

Recall that, #sinphi=sinalpha rArr phi=(-1)^kalpha+kpi, k in ZZ.#

Hence, #sin((n+1)theta)=sinntheta,#

#rArr (n+1)theta=(-1)^kntheta+kpi, k in ZZ.#

If, #k# is even, i.e., if #k=2m,# then,

#(n+1)theta=(-1)^(2m)ntheta+2mpi=ntheta+2mpi,#

#:. ((n+1)-n)theta=2mpi,#

#:. ntheta=2mpi,#

#:. theta=(2m)/npi, m in ZZ#.

Similarly, if #k# is odd, i.e., if #k=(2m+1)pi,# then

#(n+1)theta=(-1)^(2m+1)ntheta+(2m+1)pi=-ntheta+(2m+1)pi,#

#:. ((n+1)+n)theta=(2m+1)pi,#

#:. (2n+1)theta=(2m+1)pi,#

#:. theta=((2m+1)/(2n+1))pi, m in ZZ.#

Altogether, #theta in {(2m)/npi}uu{((2m+1)/(2n+1))pi}, m in ZZ.#

Enjoy Maths.!