Question

Show that

$(1 + cos \frac{\pi}{10} )(1 + cos \frac{3\pi}{10} )(1 + cos \frac{7\pi}{10} )(1 + cos \frac{9\pi}{10} ) = \frac{1}{16}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \bigg(1 + cos\dfrac{\pi}{10} \bigg)\bigg(1 + cos\dfrac{3\pi}{10} \bigg)\bigg(1 + cos\dfrac{7\pi}{10} \bigg)\bigg(1 + cos\dfrac{9\pi}{10} \bigg)$

Consider,

$\rm \: cos\bigg(\dfrac{7\pi}{10} \bigg)$

$\rm \: = \: cos\bigg(\pi - \dfrac{3\pi}{10} \bigg)$

$\rm \: = \: - \: cos\bigg(\dfrac{3\pi}{10} \bigg)$

$\rm\implies \:cos\bigg(\dfrac{7\pi}{10} \bigg) = \: - \: cos\bigg(\dfrac{3\pi}{10} \bigg)$

Now, Consider

$\rm \: cos\bigg(\dfrac{9\pi}{10} \bigg)$

$\rm \: = \: cos\bigg(\pi - \dfrac{\pi}{10} \bigg)$

$\rm \: = \: - \: cos\bigg(\dfrac{\pi}{10} \bigg)$

$\rm\implies \:cos\bigg(\dfrac{9\pi}{10} \bigg) = \: - \: cos\bigg(\dfrac{\pi}{10} \bigg)$

So, Consider again,

$\rm \: \bigg(1 + cos\dfrac{\pi}{10} \bigg)\bigg(1 + cos\dfrac{3\pi}{10} \bigg)\bigg(1 + cos\dfrac{7\pi}{10} \bigg)\bigg(1 + cos\dfrac{9\pi}{10} \bigg)$

So, this expression can be rewritten as

$\rm \: = \bigg(1 + cos\dfrac{\pi}{10} \bigg)\bigg(1 + cos\dfrac{3\pi}{10} \bigg)\bigg(1 - cos\dfrac{3\pi}{10} \bigg)\bigg(1 - cos\dfrac{\pi}{10} \bigg)$

$\rm \: = \: \bigg(1 - {cos}^{2} \dfrac{\pi}{10} \bigg)\bigg(1 - {cos}^{2} \dfrac{3\pi}{10} \bigg)$

$\rm \: = \: {sin}^{2}\bigg(\dfrac{\pi}{10} \bigg) \: {sin}^{2}\bigg(\dfrac{3\pi}{10} \bigg)$

$\rm \: = \: {sin}^{2} 18 \degree \: {sin}^{2} 54\degree$

$\rm \: = \: {(sin18\degree \times sin54\degree)}^{2}$

$\rm \: = \: {\bigg(\dfrac{ \sqrt{5} - 1}{4} \times \dfrac{ \sqrt{5} + 1}{4} \bigg)}^{2}$

$\rm \: = \: {\bigg(\dfrac{ (\sqrt{5})^{2} - 1}{16} \bigg)}^{2}$

$\rm \: = \: {\bigg(\dfrac{ 5 - 1}{16} \bigg)}^{2}$

$\rm \: = \: {\bigg(\dfrac{4}{16} \bigg)}^{2}$

$\rm \: = \: {\bigg(\dfrac{1}{4}\bigg)}^{2}$

$\rm \: = \: \dfrac{1}{16}$

Hence,

$\boxed{\sf{\bigg(1 + cos\dfrac{\pi}{10} \bigg)\bigg(1+cos\dfrac{3\pi}{10} \bigg)\bigg(1+cos\dfrac{7\pi}{10} \bigg)\bigg(1+cos\dfrac{9\pi}{10} \bigg)=\frac{1}{16}}}$

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Values Used

$\boxed{\sf{ \: {\pi}^{c} = 180\degree \: }}$

$\boxed{\sf{ \:sin18\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

$\boxed{\sf{ \:sin54\degree = \frac{ \sqrt{5} + 1}{4} \: }}$

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Additional Information

$\boxed{\sf{ \:cos36\degree = \frac{ \sqrt{5} + 1}{4} \: }}$

$\boxed{\sf{ \:cos72\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

$\boxed{\sf{ \:sin72\degree = \frac{ \sqrt{10 + 2 \sqrt{5} }}{4} \: }}$

$\boxed{\sf{ \:cos18\degree = \frac{ \sqrt{10 + 2 \sqrt{5} }}{4} \: }}$

$\boxed{\sf{ \:cos54\degree = \frac{ \sqrt{10 - 2 \sqrt{5} }}{4} \: }}$

$\boxed{\sf{ \:sin36\degree = \frac{ \sqrt{10 - 2 \sqrt{5} }}{4} \: }}$

Answer 2

Answer:

$\begin{aligned} \color{red}\text { LHS } & \color{red}=\left(1+\cos 18^{\circ}\right)\left(1+\cos 54^{\circ}\right)\left(1+\cos 126^{\circ}\right)\left(1+\cos 162^{\circ}\right) \\ \\ &\color{blue}=\left(1+\cos 18^{\circ}\right)\left(1+\cos 54^{\circ}\right)\left(1+\cos \left(180^{\circ}-54^{\circ}\right)\right) \times ( 1+ \cos(180 {}^{ \circ } - 18 {}^{ \circ} ) )\\ \\ &\color{maroon}=\left(1+\cos 18^{\circ}\right)\left(1+\cos 54^{\circ}\right)\left(1-\cos 54^{\circ}\right)\left(1-\cos 18^{\circ}\right) \\ \\ &\color{orange}=\left(1-\cos ^{2} 18^{\circ}\right)\left(1-\cos ^{2} 54^{\circ}\right) \\ \\ &\color{Crimson}=\sin ^{2} 18^{\circ} \sin ^{2} 54^{\circ}=\sin ^{2} 18^{\circ}\left(\sin \left(90^{\circ}-36^{\circ}\right)\right)^{2} \\ \\ &\color{navy}=\sin ^{2} 18^{\circ} \cos ^{2} 36^{\circ}=\left(\frac{\sqrt{5}-1}{4}\right)^{2}\left(\frac{\sqrt{5}+1}{4}\right)^{2} \\ \\ &\color{blue}=\left(\frac{5-1}{16}\right)^{2}=\frac{1}{16}=\text { RHS. } \end{aligned}$