First slide

Multiple and sub- multiple Angles

Question

The value of $\sin^{8} θ + \cos^{8} θ + \sin^{6} {θcos}^{2} θ$ $+ 3 \sin^{4} {θcos}^{2} θ + \cos^{6} {θsin}^{2} θ + 3 \sin^{2} {θcos}^{4} θ$ is equal to

Moderate

A

$\cos^{2} 2 θ$
B

$\sin^{2} 2 θ$
C

$\cos^{3} 2 θ + \sin^{3} 2 θ$
D

none of these

Solution

We have,
$\begin{aligned} \sin^{8} θ + \cos^{8} θ + \sin^{6} {θcos}^{2} θ + 3 \sin^{4} {θcos}^{2} θ + \cos^{6} {θsin}^{2} θ + 3 \sin^{2} {θcos}^{4} θ \\ = \sin^{8} θ + \cos^{8} θ + \sin^{2} {θcos}^{2} θ (\sin^{4} θ + \cos^{4} θ + 3 \sin^{2} θ + 3 \cos^{2} θ) \end{aligned}$
$\begin{aligned} = {(\sin^{4} θ + \cos^{4} θ)}^{2} - 2 \sin^{4} {θcos}^{4} θ + \frac{1}{4} \sin^{2} 2 θ {{(\sin^{2} θ + \cos^{2} θ)}^{2} - \frac{1}{2} \sin^{2} 2 θ + 3} \\ = {(1 - \frac{\sin^{2} 2 θ}{4})}^{2} - \frac{1}{8} \sin^{4} 2 θ + \sin^{2} 2 θ - \frac{1}{8} \sin^{4} 2 θ \end{aligned}$
$\begin{aligned} = 1 + \frac{1}{2} \sin^{2} 2 θ - \frac{3}{16} \sin^{4} 2 θ \\ = \frac{1}{16} (16 + 8 \sin^{2} 2 θ - 3 \sin^{4} 2 θ) \end{aligned}$

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