Answer: (i) Let \(c=\cos\theta\) and \(s=\sin\theta\). By de Moivre’s theorem,
\((c+is)^4=\cos 4\theta+i\sin 4\theta\).
Expanding the left-hand side gives
\((c+is)^4=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4\).
Using \(i^2=-1\), \(i^3=-i\), \(i^4=1\), this becomes
\((c+is)^4=(c^4-6c^2s^2+s^4)+i(4c^3s-4cs^3)\).
So
\(\cos 4\theta=c^4-6c^2s^2+s^4\) and \(\sin 4\theta=4c^3s-4cs^3\).
Hence
\(\tan 4\theta=\frac{\sin 4\theta}{\cos 4\theta}=\frac{4c^3s-4cs^3}{c^4-6c^2s^2+s^4}.\)
Now divide top and bottom by \(c^4\):
\(\tan 4\theta=\frac{4\frac{s}{c}-4\left(\frac{s}{c}\right)^3}{1-6\left(\frac{s}{c}\right)^2+\left(\frac{s}{c}\right)^4}=\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}.\)
(ii) Let \(t=\tan\theta\). If \(\tan 4\theta=-1\), then by the formula from part (i),
\(\frac{4t-4t^3}{1-6t^2+t^4}=-1.\)
Rearranging gives
\(4t-4t^3=-(1-6t^2+t^4)\)
so
\(t^4-4t^3-6t^2+4t+1=0.\)
Therefore we need \(\tan 4\theta=-1\). The angles with tangent \(-1\) are
\(4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4},\ \frac{11\pi}{4},\ \frac{15\pi}{4}\)
modulo \(2\pi\). Dividing by 4 gives
\(\theta=\frac{3\pi}{16},\ \frac{7\pi}{16},\ \frac{11\pi}{16},\ \frac{15\pi}{16}.\)
So the solutions are
\(t=\tan\frac{3\pi}{16},\ \tan\frac{7\pi}{16},\ \tan\frac{11\pi}{16},\ \tan\frac{15\pi}{16}.\)
(i) Set \(c=\cos\theta\) and \(s=\sin\theta\). Then by de Moivre’s theorem
\((c+is)^4=\cos 4\theta+i\sin 4\theta.\)
Expand the left-hand side:
\((c+is)^4=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4.\)
Now \((is)^2=-s^2\), \((is)^3=-is^3\), and \((is)^4=s^4\), so
\((c+is)^4=(c^4-6c^2s^2+s^4)+i(4c^3s-4cs^3).\)
Equating real and imaginary parts gives
\(\cos 4\theta=c^4-6c^2s^2+s^4,\)
\(\sin 4\theta=4c^3s-4cs^3.\)
Hence
\(\tan 4\theta=\frac{\sin 4\theta}{\cos 4\theta}=\frac{4c^3s-4cs^3}{c^4-6c^2s^2+s^4}.\)
Dividing numerator and denominator by \(c^4\) and writing \(t=\tan\theta=\frac{s}{c}\),
\(\tan 4\theta=\frac{4t-4t^3}{1-6t^2+t^4}.\)
This is the required result.
(ii) Let \(t\) be a root of
\(t^4-4t^3-6t^2+4t+1=0.\)
Rearranging,
\(4t-4t^3=-(1-6t^2+t^4),\)
so
\(\frac{4t-4t^3}{1-6t^2+t^4}=-1.\)
Using part (i), this means \(\tan 4\theta=-1\) when \(t=\tan\theta\). Therefore
\(4\theta=\frac{3\pi}{4}+n\pi\)
for some integer \(n\). Taking one complete set of values modulo \(2\pi\),
\(4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4},\ \frac{11\pi}{4},\ \frac{15\pi}{4}.\)
Dividing by 4 gives
\(\theta=\frac{3\pi}{16},\ \frac{7\pi}{16},\ \frac{11\pi}{16},\ \frac{15\pi}{16}.\)
Hence the solutions are
\(t=\tan\frac{3\pi}{16},\ \tan\frac{7\pi}{16},\ \tan\frac{11\pi}{16},\ \tan\frac{15\pi}{16}.\)
