Question

(1+cos π÷10)(1+cos 3π÷10)(1+cos 7π÷10)(1+cos 9π÷10)=1÷16

Answer 1

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

Consider,

$\rm :\longmapsto\:\bigg[1 + cos\dfrac{\pi}{10} \bigg]\bigg[1 + cos\dfrac{3\pi}{10} \bigg]\bigg[1 + cos\dfrac{7\pi}{10} \bigg]\bigg[1 + cos\dfrac{9\pi}{10} \bigg]$

Now,

We know that,

$\boxed{ \tt{ \: cos[\pi - x] = - \: cosx}}$

So, Consider,

$\rm :\longmapsto\:cos\bigg[\dfrac{7\pi}{10} \bigg] = cos\bigg[\pi - \dfrac{3\pi}{10} \bigg] = - \: cos\bigg[\dfrac{3\pi}{10} \bigg]$

and

$\rm :\longmapsto\:cos\bigg[\dfrac{9\pi}{10} \bigg] = cos\bigg[\pi - \dfrac{\pi}{10} \bigg] = - \: cos\bigg[\dfrac{\pi}{10} \bigg]$

So, on substituting these values in given expression, we get

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

can be re-arranged as

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

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

We know,

$\purple{\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}}$

So, using this, we get

$\rm \: = \: {sin}^{2}\bigg[\dfrac{\pi}{10} \bigg] \times {sin}^{2} \bigg[\dfrac{3\pi}{10} \bigg]$

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

$\rm \: = \: {[sin18 \degree \times sin54\degree]}^{2}$

We know,

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

and

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

So, using these values, we get

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

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

$\rm \: = \: {\bigg[\dfrac{ 5 - 1}{16} \bigg]}^{2}$

$\rm \: = \: {\bigg[\dfrac{4}{16} \bigg]}^{2}$

$\rm \: = \: {\bigg[\dfrac{1}{4} \bigg]}^{2}$

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

Hence,

$\red{\boxed{ \sf{ \: \bigg[1 + cos\dfrac{\pi}{10} \bigg]\bigg[1 + cos\dfrac{3\pi}{10} \bigg]\bigg[1 + cos\dfrac{7\pi}{10} \bigg]\bigg[1 + cos\dfrac{9\pi}{10} \bigg] = \dfrac{1}{16} }}}$

Additional Information :-

$\boxed{ \tt{ \: sin18\degree = cos72\degree = \frac{ \sqrt{5} - 1 }{4}}}$

$\boxed{ \tt{ \: sin54\degree = cos36\degree = \frac{ \sqrt{5} - 1 }{4}}}$

$\boxed{ \tt{ \: cos18\degree = sin72\degree = \sqrt{\dfrac{10 + 2 \sqrt{5} }{4} }}}$

$\boxed{ \tt{ \: cos54\degree = sin36\degree = \sqrt{\dfrac{10 - 2 \sqrt{5} }{4} }}}$