Question 1
Show that \( \cos 4 \theta = 8 \cos^4 \theta-8 \cos^2 \theta + 1 \).
(a) By considering the real part of \( z^4 \), prove \( \cos 4\theta = \cos^4 \theta-6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta \) by letting \( z = \cos \theta + i \sin \theta \).
\( \require{AMSsymbols} \begin{align} \displaystyle \require{color}
z &= \cos \theta + i \sin \theta \\
z^4 &= (\cos \theta + i \sin \theta)^4 \\
& = \cos 4\theta + i \sin 4\theta \cdots (1) &\color{red} \text{by De Moivre’s theorem}\\
z^4 &= (\cos \theta + i \sin \theta)^4 \\
&= \cos^4 \theta + 4 i \cos^3\theta \sin \theta + 6 i^2 \cos^2 \theta \sin ^2 \theta + 4i \cos \theta \sin^3 \theta + i^4 \sin \theta \\
&= \cos^4 \theta + 4 i \cos^3\theta \sin \theta-6 \cos^2 \theta \sin ^2 \theta + 4i \cos \theta \sin^3 \theta + \sin^4 \theta \\
&= (\cos^4 \theta-6 \cos^2 \theta \sin ^2 \theta + \sin^4 \theta ) + i(4 \cos^3\theta \sin \theta + 4 \cos \theta \sin^3 \theta) \cdots (2) \\
\end{align} \)
Equate only the real component of the expansion of \( (1) \) and \( (2) \).
\( \therefore \cos 4\theta = \cos^4 \theta-6 \cos^2 \theta \sin ^2 \theta + \sin^4 \theta \)
(b) Hence, find an expression for \( \sin 4\theta \) involving powers of \( \sin \theta \) and \( \cos \theta \).
\( \require{AMSsymbols} \begin{align} \require{color}
\sin 4\theta &= 4 \cos^3\theta \sin \theta + 4 \cos \theta \sin^3 \theta &\color{red} \text{equating the imaginery parts}\\
\end{align} \)
(c) Hence, find an expression for \( \cos 4\theta \) involving only powers of \( \cos \theta \).
\( \require{AMSsymbols} \begin{align} \require{color}
\cos 4\theta &= \cos^4 \theta-6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta &\color{red} \text{equating the real parts}\\
&= \cos^4 \theta-6 \cos^2 \theta (1-\cos^2 \theta) + (1-\cos^2 \theta)^2 \\
&= \cos^4 \theta-6 \cos^2 \theta + 6 \cos^4 \theta + 1-2 \cos^2 \theta + \cos^4 \theta \\
&= 8 \cos^4 \theta-8 \cos^2 \theta + 1
\end{align} \)
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