Answer: (a) \(\displaystyle \cos^7\theta=\frac{1}{64}\cos 7\theta+\frac{7}{64}\cos 5\theta+\frac{21}{64}\cos 3\theta+\frac{35}{64}\cos\theta\).

Hence \(a=\frac{1}{64},\ b=\frac{7}{64},\ c=\frac{21}{64},\ d=\frac{35}{64}\).

(b) \(\displaystyle nI_n=2^{-n/2}+(n-1)I_{n-2}\).

(c) \(\displaystyle I_9=\frac{2867}{10080}\sqrt2\).

(a) Let \(z=\cos\theta+\mathrm{i}\sin\theta\). Then \(z^{-1}=\cos\theta-\mathrm{i}\sin\theta\), so

\(z+z^{-1}=2\cos\theta.\)

Therefore

\((z+z^{-1})^7=(2\cos\theta)^7=2^7\cos^7\theta.\)

Expanding and grouping symmetric terms,

\((z+z^{-1})^7=(z^7+z^{-7})+7(z^5+z^{-5})+21(z^3+z^{-3})+35(z+z^{-1}).\)

By de Moivre's theorem, \(z^n+z^{-n}=2\cos n\theta\). Hence

\(2^7\cos^7\theta=2\cos7\theta+7(2\cos5\theta)+21(2\cos3\theta)+35(2\cos\theta).\)

Dividing by \(128\),

\(\cos^7\theta=\frac{1}{64}\cos7\theta+\frac{7}{64}\cos5\theta+\frac{21}{64}\cos3\theta+\frac{35}{64}\cos\theta.\)

So \(a=\frac{1}{64},\ b=\frac{7}{64},\ c=\frac{21}{64},\ d=\frac{35}{64}\).

(b) Start from

\(I_n=\int_0^{\pi/4}\cos^{n-1}\theta\cos\theta\,\mathrm d\theta.\)

Use integration by parts with \(u=\cos^{n-1}\theta\), \(\mathrm dv=\cos\theta\,\mathrm d\theta\). Then

\(\mathrm du=-(n-1)\cos^{n-2}\theta\sin\theta\,\mathrm d\theta,\quad v=\sin\theta.\)

So

\(I_n=\left[\cos^{n-1}\theta\sin\theta\right]_0^{\pi/4}+(n-1)\int_0^{\pi/4}\cos^{n-2}\theta\sin^2\theta\,\mathrm d\theta.\)

Using \(\sin^2\theta=1-\cos^2\theta\),

\(I_n=\left[\cos^{n-1}\theta\sin\theta\right]_0^{\pi/4}+(n-1)\int_0^{\pi/4}\cos^{n-2}\theta(1-\cos^2\theta)\,\mathrm d\theta.\)

Hence

\(I_n=\left[\cos^{n-1}\theta\sin\theta\right]_0^{\pi/4}+(n-1)I_{n-2}-(n-1)I_n.\)

Now

\(\left[\cos^{n-1}\theta\sin\theta\right]_0^{\pi/4}=\left(\frac{1}{\sqrt2}\right)^{n-1}\left(\frac{1}{\sqrt2}\right)=2^{-n/2}.\)

Therefore

\(I_n=2^{-n/2}+(n-1)I_{n-2}-(n-1)I_n,\)

so

\(nI_n=2^{-n/2}+(n-1)I_{n-2}.\)

(c) Apply the reduction formula with \(n=9\):

\(9I_9=2^{-9/2}+8I_7=\frac{1}{16}\,2^{-1/2}+8I_7.\)

From part (a),

\(\cos^7\theta=\frac{1}{64}\cos7\theta+\frac{7}{64}\cos5\theta+\frac{21}{64}\cos3\theta+\frac{35}{64}\cos\theta.\)

So

\(I_7=\int_0^{\pi/4}\cos^7\theta\,\mathrm d\theta\)

\(=\frac{1}{64}\left[\frac{1}{7}\sin7\theta+\frac{7}{5}\sin5\theta+\frac{21}{3}\sin3\theta+35\sin\theta\right]_0^{\pi/4}.\)

Using \(\sin\frac{7\pi}{4}=-\frac{1}{\sqrt2}\), \(\sin\frac{5\pi}{4}=-\frac{1}{\sqrt2}\), \(\sin\frac{3\pi}{4}=\frac{1}{\sqrt2}\), \(\sin\frac{\pi}{4}=\frac{1}{\sqrt2}\), we get

\(I_7=\frac{1}{64\sqrt2}\left(-\frac{1}{7}-\frac{7}{5}+7+35\right)=\frac{177}{280\sqrt2}.\)

Hence

\(9I_9=\frac{1}{16\sqrt2}+8\cdot\frac{177}{280\sqrt2}=\frac{1}{16\sqrt2}+\frac{177}{35\sqrt2}.\)

With common denominator,

\(9I_9=\frac{2867}{560\sqrt2}.\)

Therefore

\(I_9=\frac{2867}{5040\sqrt2}=\frac{2867}{10080}\sqrt2.\)