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Question Number 117257 by bemath last updated on 10/Oct/20

If sin^2 (x)+cos^2 (x)=1 then   what the value of sin^(11) (x)+cos^(11) (x) =?

Ifsin2(x)+cos2(x)=1thenwhatthevalueofsin11(x)+cos11(x)=?

Commented by bemath last updated on 10/Oct/20

nice question

nicequestion

Answered by mathmax by abdo last updated on 10/Oct/20

generally let find sin^(2n+1) x +cos^(2n+1) x =u_n (x)  u_n (x) = (((e^(ix) −e^(−ix) )/(2i)))^(2n+1)  +(((e^(ix) +e^(−ix) )/2))^(2n+1)   =(1/((2i)^(2n+1) ))Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  (e^(ix) )^k (−e^(−ix) )^(2n+1−k)   +(1/2^(2n+1) )Σ_(k=0) ^(2n+1) C_(2n+1) ^k  (e^(ix) )^k (e^(−ix) )^(2n+1−k)   =(1/((2i)^(2n+1) ))Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(ikx) (−1)^(2n+1−k)  e^(−i(2n+1−k)x)   +(1/2^(2n+1) ) Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(ikx)  e^(−i(2n+1−k)x)   =−(((−1)^n )/(2^(2n+1) i)) Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(i(k−(2n+1)+k)x)   +(1/2^(2n+1) ) Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(i(k+k−(2n+1))x)   =i (((−1)^n )/2^(2n+1) ) Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(i(2k−(2n+1))x)  +(1/2^(2n+1) )Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(i(2k−(2n+1))x)   =(1+i(−1)^n )×(1/2^(2n+1) ) Σ_(k=0) ^(2n+1)  C_(2n+1) ^k  e^(i(2k−(2n+1))x)   n=5 ⇒sin^(11) x +cos^(11) x =((1−i)/2^(11) ) Σ_(k=0) ^(11)  C_(11) ^k  e^(i(2k−11)x)   =((1−i)/2^(11) ){ C_(11) ^0  e^(−11ix)  +C_(11) ^1  e^(i(−9)x)  +C_(11) ^2  e^(i(−7)x) +....C_(11) ^(11)  e^(i11x) }

generallyletfindsin2n+1x+cos2n+1x=un(x)un(x)=(eixeix2i)2n+1+(eix+eix2)2n+1=1(2i)2n+1k=02n+1C2n+1k(eix)k(eix)2n+1k+122n+1k=02n+1C2n+1k(eix)k(eix)2n+1k=1(2i)2n+1k=02n+1C2n+1keikx(1)2n+1kei(2n+1k)x+122n+1k=02n+1C2n+1keikxei(2n+1k)x=(1)n22n+1ik=02n+1C2n+1kei(k(2n+1)+k)x+122n+1k=02n+1C2n+1kei(k+k(2n+1))x=i(1)n22n+1k=02n+1C2n+1kei(2k(2n+1))x+122n+1k=02n+1C2n+1kei(2k(2n+1))x=(1+i(1)n)×122n+1k=02n+1C2n+1kei(2k(2n+1))xn=5sin11x+cos11x=1i211k=011C11kei(2k11)x=1i211{C110e11ix+C111ei(9)x+C112ei(7)x+....C1111ei11x}

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