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Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \frac{2 - (x^2 - 1)^2}{2}$.
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Given: $\sin \theta + \cos \theta = x$.
Squaring both sides
$\sin^2 \theta + \cos^2 \theta + 2\cos \theta \sin \theta = x^2$. $2\sin \theta \cos \theta = x^2 - 1$ (1 mark).
$RHS = \frac{2 - (2\sin \theta \cos \theta)^2}{2} = \frac{2 - 4\sin^2 \theta \cos^2 \theta}{2} = 1 - 2\sin^2 \theta \cos^2 \theta$ ($\frac{1}{2} + \frac{1}{2}$ marks).
$$\begin{aligned}& = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = (\sin^4 \theta + \cos^4 \theta) \\ & = LHS\end{aligned}$$ ($\frac{1}{2} + \frac{1}{2}$ marks).
Squaring both sides
$\sin^2 \theta + \cos^2 \theta + 2\cos \theta \sin \theta = x^2$. $2\sin \theta \cos \theta = x^2 - 1$ (1 mark).
$RHS = \frac{2 - (2\sin \theta \cos \theta)^2}{2} = \frac{2 - 4\sin^2 \theta \cos^2 \theta}{2} = 1 - 2\sin^2 \theta \cos^2 \theta$ ($\frac{1}{2} + \frac{1}{2}$ marks).
$$\begin{aligned}& = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = (\sin^4 \theta + \cos^4 \theta) \\ & = LHS\end{aligned}$$ ($\frac{1}{2} + \frac{1}{2}$ marks).