Question #117392
Express sin 5θ and cos 5θ/ cos θ in terms of sin θ
1
Expert's answer
2020-05-25T20:56:15-0400

sin(5θ)=sin(2θ+3θ)=sin(2θ)cos(3θ)+sin(3θ)cos(2θ)sin(3θ)=3sin(θ)cos2(θ)sin3(θ)cos(3θ)=cos3(θ)3cos(θ)sin2(θ)sin(2θ)=2sin(θ)cos(θ)cos(2θ)=1sin2(θ)sin(5θ)=2sin(θ)cos(θ)×(cos3(θ)3cos(θ)sin2(θ))++(3sin(θ)cos2(θ)sin3(θ))×(1sin2(θ))==2sin(θ)cos4(θ)6sin3(θ)cos2(θ)+3sin(θ)cos2(θ)3sin3(θ)cos2(θ)sin3(θ)+sin5(θ)==2sin(θ)(1sin2(θ))26sin3(θ)(1sin2(θ))+3sin(θ)(1sin2(θ))3sin3(θ)(1sin2(θ))sin3(θ)+sin5(θ)==2sin(θ)4sin3(θ)+2sin5(θ)6sin3(θ)+6sin5(θ)+3sin(θ)3sin3(θ)3sin3(θ)+3sin5(θ)sin3(θ)+sin5(θ)==13sin5(θ)17sin3(θ)+5sin(θ)\sin(5\theta) = \sin(2\theta+3\theta)=\sin(2\theta)\cos(3\theta)+\sin(3\theta)cos(2\theta)\newline \sin(3\theta) = 3\sin(\theta)\cos^2(\theta) – \sin^3(\theta)\newline \cos(3\theta) = \cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)\newline \sin(2\theta) = 2\sin(\theta)\cos(\theta)\newline \cos(2\theta)=1-\sin^2(\theta)\newline \sin(5\theta)= 2\sin(\theta)\cos(\theta) \times (\cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)) +\newline +(3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))\times(1-\sin^2(\theta))=\newline =2\sin(\theta)\cos^4(\theta)-6\sin^3(\theta)\cos^2(\theta)+3\sin(\theta)\cos^2(\theta)-\newline -3\sin^3(\theta)cos^2(\theta)-\sin^3(\theta)+\sin^5(\theta)=\newline =2\sin(\theta)(1-\sin^2(\theta))^2-6\sin^3(\theta)(1-\sin^2(\theta))+3\sin(\theta)(1-\sin^2(\theta))-\newline -3\sin^3(\theta)(1-\sin^2(\theta))-\sin^3(\theta)+\sin^5(\theta)=\newline =2\sin(\theta)-4\sin^3(\theta)+2\sin^5(\theta)-6\sin^3(\theta)+6\sin^5(\theta)+3\sin(\theta)-3\sin^3(\theta)-\newline -3\sin^3(\theta)+3\sin^5(\theta)-\sin^3(\theta)+\sin^5(\theta)=\newline =13\sin^5(\theta)-17\sin^3(\theta)+5\sin(\theta)\newline


cos(5θ)=cos(3θ+2θ)=cos(3θ)cos(2θ)sin(3θ)sin(2θ)\cos(5\theta)=\cos(3\theta+2\theta)=\cos(3\theta)\cos(2\theta)-\sin(3\theta)\sin(2\theta)


From the previous paragraph:

sin(3θ)=3sin(θ)cos2(θ)sin3(θ)cos(3θ)=cos3(θ)3cos(θ)sin2(θ)sin(2θ)=2sin(θ)cos(θ)cos(2θ)=1sin2(θ)So:cos(5θ)=(cos3(θ)3cos(θ)sin2(θ))(1sin2(θ))(3sin(θ)cos2(θ)sin3(θ))×2sin(θ)cos(θ))==cos(θ)(cos2(θ)3sin2(θ))(1sin2(θ))2sin(θ)cos(θ)(3sin(θ)cos2(θ)sin3(θ))==cos(θ)((14sin2(θ))(1sin2(θ))2sin(θ)(3sin(θ)4sin3(θ))==cos(θ)(15sin2(θ)+4sin4(θ)6sin2(θ)+8sin4(θ))==cos(θ)(111sin2(θ)+12sin4(θ))=>cos(5θ)cos(θ)=12sin4(θ)11sin2(θ)+1\sin(3\theta) = 3\sin(\theta)\cos^2(\theta) – \sin^3(\theta)\newline \cos(3\theta) = \cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)\newline \sin(2\theta) = 2\sin(\theta)\cos(\theta)\newline \cos(2\theta)=1-\sin^2(\theta)\newline So:\newline \cos(5\theta)= (\cos^3(\theta) – 3\cos(\theta)\sin^2(\theta))(1-\sin^2(\theta))-\newline - (3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))\times2\sin(\theta)\cos(\theta))=\newline =\cos(\theta)(\cos^2(\theta) – 3\sin^2(\theta))(1-\sin^2(\theta))-\newline -2\sin(\theta)\cos(\theta)(3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))=\newline =\cos(\theta)((1-4\sin^2(\theta))(1-\sin^2(\theta))-2\sin(\theta)(3\sin(\theta)-4\sin^3(\theta))=\newline =\cos(\theta)(1-5\sin^2(\theta)+4\sin^4(\theta)-6\sin^2(\theta)+8\sin^4(\theta))=\newline =\cos(\theta)(1-11\sin^2(\theta)+12\sin^4(\theta))=>\newline \dfrac{\cos(5\theta)}{\cos(\theta)}=12\sin^4(\theta)-11\sin^2(\theta)+1



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