Answer: Either

Using \(1+i\tan\theta)^k=\sec^k\theta\, (\cos k\theta+i\sin k\theta)\), we have

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\sum_{k=0}^{n-1}\sec^k\theta\,(\cos k\theta+i\sin k\theta)\).

Now the sum of a geometric progression gives

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\dfrac{(1+i\tan\theta)^n-1}{i\tan\theta}\).

Also \(1+i\tan\theta=\sec\theta(\cos\theta+i\sin\theta)\), so

\((1+i\tan\theta)^n=\sec^n\theta\, (\cos n\theta+i\sin n\theta)\).

Hence

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\dfrac{\sec^n\theta(\cos n\theta+i\sin n\theta)-1}{i\tan\theta}\).

Separating real and imaginary parts and using \(\cot\theta=1/\tan\theta\), the real part gives

\(\sum_{k=0}^{n-1}\cos k\theta\,\sec^k\theta=\sec^n\theta\,\sin n\theta\,\cot\theta\).

Therefore

\(\sum_{k=0}^{n-1}\cos k\theta\,\sec^k\theta=\cot\theta\,\sin n\theta\,\sec^n\theta\),

as required.

Now put \(\theta=\pi/3\). Then \(\sec(\pi/3)=2\) and \(\cot(\pi/3)=1/\sqrt{3}\), so

\(\sum_{k=0}^{n-1}2^k\cos\left(\frac{k\pi}{3}\right)=\frac{2^n}{\sqrt{3}}\sin\left(\frac{n\pi}{3}\right)\).

Finally, if \(0\lt x\lt 1\) and \(\theta=\cos^{-1}x\), then \(x=\cos\theta\), so \(\sec\theta=1/x\) and

\(\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{x}{\sqrt{1-x^2}}\).

Substituting into the first identity gives

\(\sum_{k=0}^{n-1}\dfrac{\cos(k\cos^{-1}x)}{x^k}=\dfrac{\sin(n\cos^{-1}x)}{x^{n-1}\sqrt{1-x^2}}\).

Either

For each integer \(k\), write

\(1+i\tan\theta=\sec\theta(\cos\theta+i\sin\theta)\).

Therefore

\((1+i\tan\theta)^k=\sec^k\theta\,(\cos k\theta+i\sin k\theta)\).

So

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\sum_{k=0}^{n-1}\sec^k\theta\,(\cos k\theta+i\sin k\theta)\).

Because this is a geometric series,

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\dfrac{(1+i\tan\theta)^n-1}{i\tan\theta}\).

Now expand the numerator using the identity for \((1+i\tan\theta)^n\):

\((1+i\tan\theta)^n=\sec^n\theta\,(\cos n\theta+i\sin n\theta)\).

Hence

\(\sum_{k=0}^{n-1}(1+i\tan\theta)^k=\dfrac{\sec^n\theta(\cos n\theta+i\sin n\theta)-1}{i\tan\theta}\).

After simplifying, the real part is

\(\sec^n\theta\,\sin n\theta\,\cot\theta\).

But the real part of the left-hand side is exactly \(\sum_{k=0}^{n-1}\cos k\theta\,\sec^k\theta\). Therefore

\(\sum_{k=0}^{n-1}\cos k\theta\,\sec^k\theta=\cot\theta\,\sin n\theta\,\sec^n\theta\).

This proves the first result.

To obtain the second result, take \(\theta=\pi/3\). Then \(\sec\theta=2\) and \(\cot\theta=1/\sqrt{3}\), so

\(\sum_{k=0}^{n-1}2^k\cos\left(\frac{k\pi}{3}\right)=\frac{2^n}{\sqrt{3}}\sin\left(\frac{n\pi}{3}\right)\).

For the final identity, let \(\theta=\cos^{-1}x\). Since \(0\lt x\lt 1\), we have \(0\lt \theta\lt \pi/2\), so

\(\sec\theta=\frac{1}{x}\), \(\sin\theta=\sqrt{1-x^2}\), and thus

\(\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{x}{\sqrt{1-x^2}}\).

Substituting these into the first identity gives

\(\sum_{k=0}^{n-1}\frac{\cos(k\cos^{-1}x)}{x^k}=\frac{1}{\sqrt{1-x^2}}\cdot \sin\bigl(n\cos^{-1}x\bigr)\cdot \frac{1}{x^{n-1}}\).

So

\(\sum_{k=0}^{n-1}\dfrac{\cos\left(k\cos^{-1}x\right)}{x^k}=\dfrac{\sin\left(n\cos^{-1}x\right)}{x^{n-1}\sqrt{1-x^2}}\).