Question

Cos inverse (a+bcosx/b+acosx)

Answer 1

Answer:

don't know sorry.....

Answer 2

Answer :

arccos(b+acosxa+bcosx)udFdx=arccos(u)=F(u(x))=b+acosxa+bcosx=dFdududx

arccos⁡(b+acos⁡xa+bcos⁡x)=arccos⁡(u)=F(u(x))u=b+acos⁡xa+bcos⁡xdFdx=dFdududx

dFdu=−11−u2−−−−−√=−11−(b+acosxa+bcosx)2−−−−−−−−−−−−√

dFdu=−11−u2=−11−(b+acos⁡xa+bcos⁡x)2

dudx=(−asinx)(a+bcosx)−(b+acosx)(−bsinx)(a+bcosx)2=1(a+bcosx)2[−a2sinx−absinxcosx+b2sinx+absinxcosx]=1(a+bcosx)2[−a2sinx+b2sinx]=(b2−a2)sinx(a+bcosx)2

dudx=(−asin⁡x)(a+bcos⁡x)−(b+acos⁡x)(−bsin⁡x)(a+bcos⁡x)2=1(a+bcos⁡x)2[−a2sin⁡x−absin⁡xcos⁡x+b2sinx+absin⁡xcos⁡x]=1(a+bcos⁡x)2[−a2sin⁡x+b2sinx]=(b2−a2)sin⁡x(a+bcos⁡x)2

dFdx=dFdududx=−11−(b+acosxa+bcosx)2−−−−−−−−−−−−√(b2−a2)sinx(a+bcosx)2=−1(a+bcosx)2−(b+acosx)2−−−−−−−−−−−−−−−−−−−−−−−√(b2−a2)sinx|a+bcosx|=sgn(sinx)a2−b2−−−−−−√|a+bcosx|