Question

$\sf{If \ sin\theta+sin^{2}\theta+sin^{3}\theta=1,}$
$\sf{then \ cos^{6}\theta-4cos^{4}\theta+8cos^{2}\theta=.....}$​

Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\rm \to \: \sin( \theta) \: + \sin {}^{2} ( \theta) + \sin {}^{3} ( \theta) = 1$

$\bf \to \: \orange{{ \underline{to \: find \div }}}$

$\cos {}^{6} ( \theta) - 4 \cos {}^{4} ( \theta) + 8 \cos {}^{2} ( \theta) =$

$\bf \to \: \blue{{ \underline{explanation \div }}}$

$\rm \to \: \sin( \theta) + \sin {}^{2} ( \theta) + \sin {}^{3} ( \theta) = 1 \\ \\ \rm \to \: \sin( \theta) + \sin {}^{3} ( \theta) = 1 - \sin {}^{2} ( \theta) \\ \\ \rm \to \: \sin( \theta) (1 \: + \sin {}^{2} ( \theta) ) = \cos {}^{2} ( \theta) \\ \\ \rm \to \: \sin( \theta) (1 + 1 - \cos {}^{2} ( \theta) ) = \cos {}^{2} ( \theta) \\ \\ \rm \to \: \sin( \theta) (2 - \cos {}^{2} ( \theta) ) = \cos {}^{2} ( \theta) \\ \\ \rm \to \: \: squaring \: on \: both \: sides \\ \\ \rm \to \: \sin {}^{2} ( \theta) (2 - \cos {}^{2} ( \theta)) {}^{2} = \cos {}^{2} ( \theta) {}^{2} \\ \\ \rm \to \: (1 - \cos {}^{2} ( \theta) )(4 \: + \: \cos {}^{4} ( \theta) - 4 \cos {}^{2} ( \theta) ) = \cos {}^{4} ( \theta) \\ \\ \rm \to \: 4 \: + \: \cos {}^{4} ( \theta) - 4 \cos {}^{2} ( \theta) - 4 \cos {}^{2} ( \theta) - \cos {}^{6} ( \theta) + \: 4 \cos {}^{4} ( \theta) = \cos {}^{4} ( \theta) \\ \\ \rm \to \: 4 \: - \: 8 \cos {}^{2} ( \theta) - \: \cos {}^{6} ( \theta) + \: 4 \cos {}^{4} ( \theta) = 0 \\ \\ \rm \to \: 4 = \cos {}^{6} ( \theta) - 4 \cos {}^{4} ( \theta) + 8 \cos {}^{2} ( \theta)$

Answer 2

Question

$\rm \: if \: \: \sin \theta + \sin {}^{2} \theta + \sin {}^{3} \theta = 1$

$\rm \: then \: find \: the \: value \: of \\ \rm \cos {}^{6} \theta - 4 \cos {}^{4} \theta + 8 \cos {}^{2} \theta$

Solution:-

Take

$\rm \: \: \: \sin( \theta) + \sin {}^{2} ( \theta) + \sin {}^{3} ( \theta) = 1$

$\rm \: \: \: \sin( \theta) + \sin {}^{3} ( \theta) = 1 - \sin {}^{2} \theta$

Use this identity

$\rm \: 1 - \sin {}^{2} \theta= \cos {}^{2} \theta$

We get

$\rm \: \: \: \sin \theta + \sin {}^{3} \theta = \cos {}^{2} \theta$

$\rm \: \sin\theta(1 + \sin {}^{2} \theta ) = \cos {}^{2} \theta$

Squaring on both side

$\rm \: \sin {}^{2} (1 + \sin {}^{2} \theta) {}^{2} = \cos {}^{4} \theta$

$\rm(1 - \cos {}^{2} \theta )(1 + 1 - \cos {}^{2} \theta) {}^{2} = \cos {}^{2} ( \theta)$

$\rm \: (1 - \cos {}^{2} \theta )(2 - \cos {}^{2} \theta ) {}^{2} = \cos {}^{4} \theta$

Use this identity

$\rm(a - b) { }^{2} = { a}^{2} + b {}^{2} - 2ab$

we get

$\rm(1 - \cos {}^{2} \theta)(4 + \cos {}^{4} \theta - 4 \cos {}^{2} \theta) = \cos {}^{4} \theta$

Now multiply

$\rm4 + \cos {}^{4} \theta - 4 \cos {}^{2} \theta - 4 \cos {}^{2} \theta \\ - \cos {}^{6} \theta+ 4 \cos {}^{4} \theta = \cos {}^{4} \theta$

we get

$\rm \: - \cos {}^{6} \theta+ 4 \cos {}^{4} \theta - 8 \cos {}^{2} \theta + 4 = 0$

$\rm \ \cos {}^{6} \theta- 4 \cos {}^{4} \theta + 8 \cos {}^{2} \theta = 4$

Answer is 4