Question

Express sin 5θ and cos 5θ/ cos θ in terms of sin θ

Answer 1

$\sin(5\theta) = \sin(2\theta+3\theta)=\sin(2\theta)\cos(3\theta)+\sin(3\theta)cos(2\theta)\newline \sin(3\theta) = 3\sin(\theta)\cos^2(\theta) – \sin^3(\theta)\newline \cos(3\theta) = \cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)\newline \sin(2\theta) = 2\sin(\theta)\cos(\theta)\newline \cos(2\theta)=1-\sin^2(\theta)\newline \sin(5\theta)= 2\sin(\theta)\cos(\theta) \times (\cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)) +\newline +(3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))\times(1-\sin^2(\theta))=\newline =2\sin(\theta)\cos^4(\theta)-6\sin^3(\theta)\cos^2(\theta)+3\sin(\theta)\cos^2(\theta)-\newline -3\sin^3(\theta)cos^2(\theta)-\sin^3(\theta)+\sin^5(\theta)=\newline =2\sin(\theta)(1-\sin^2(\theta))^2-6\sin^3(\theta)(1-\sin^2(\theta))+3\sin(\theta)(1-\sin^2(\theta))-\newline -3\sin^3(\theta)(1-\sin^2(\theta))-\sin^3(\theta)+\sin^5(\theta)=\newline =2\sin(\theta)-4\sin^3(\theta)+2\sin^5(\theta)-6\sin^3(\theta)+6\sin^5(\theta)+3\sin(\theta)-3\sin^3(\theta)-\newline -3\sin^3(\theta)+3\sin^5(\theta)-\sin^3(\theta)+\sin^5(\theta)=\newline =13\sin^5(\theta)-17\sin^3(\theta)+5\sin(\theta)\newline$

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

From the previous paragraph:

$\sin(3\theta) = 3\sin(\theta)\cos^2(\theta) – \sin^3(\theta)\newline \cos(3\theta) = \cos^3(\theta) – 3\cos(\theta)\sin^2(\theta)\newline \sin(2\theta) = 2\sin(\theta)\cos(\theta)\newline \cos(2\theta)=1-\sin^2(\theta)\newline$

So:

$\cos(5\theta)= (\cos^3(\theta) – 3\cos(\theta)\sin^2(\theta))(1-\sin^2(\theta))-\newline - (3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))\times2\sin(\theta)\cos(\theta))=\newline =\cos(\theta)(\cos^2(\theta) – 3\sin^2(\theta))(1-\sin^2(\theta))-\newline -2\sin(\theta)\cos(\theta)(3\sin(\theta)\cos^2(\theta) – \sin^3(\theta))=\newline =\cos(\theta)((1-4\sin^2(\theta))(1-\sin^2(\theta))-2\sin(\theta)(3\sin(\theta)-4\sin^3(\theta))=\newline =\cos(\theta)(1-5\sin^2(\theta)+4\sin^4(\theta)-6\sin^2(\theta)+8\sin^4(\theta))=\newline =\cos(\theta)(1-11\sin^2(\theta)+12\sin^4(\theta))=>\newline \dfrac{\cos(5\theta)}{\cos(\theta)}=12\sin^4(\theta)-11\sin^2(\theta)+1$