Question

The value of cos³(π/5) + cos³(2π/5) + cos³(3π/5) + cos³(4π/5) is ..



a) 0
b) 1
c) -1
d) None​

Answer 1

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

Given Trigonometric function is

$\rm :\longmapsto\: {cos}^{3}\bigg[\dfrac{\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{2\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{3\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{4\pi}{5} \bigg]$

Now,

Consider,

$\red{\rm :\longmapsto\:{cos}^{}\bigg[\dfrac{3\pi}{5} \bigg]}$

$\red{ \rm \: = \: {cos}^{}\bigg[\pi - \dfrac{2\pi}{5} \bigg]}$

We know,

$\boxed{ \rm \:cos(\pi - x) = \: - \: cosx}$

So, using this we get

$\red{ \rm \: = - \: {cos}^{}\bigg[ \dfrac{2\pi}{5} \bigg]}$

Hence,

$\red{\rm :\longmapsto\:{cos}^{}\bigg[\dfrac{3\pi}{5} \bigg] = - \: {cos}^{}\bigg[\dfrac{2\pi}{5} \bigg]}$

Now, Consider

$\green{\rm :\longmapsto\:{cos}^{}\bigg[\dfrac{4\pi}{5} \bigg]}$

$\green{ \rm \: = \: {cos}^{}\bigg[\pi - \dfrac{\pi}{5} \bigg]}$

$\green{ \rm \: = - \: {cos}^{}\bigg[ \dfrac{\pi}{5} \bigg]}$

Hence,

$\green{\rm :\longmapsto\:{cos}^{}\bigg[\dfrac{4\pi}{5} \bigg] = - \: {cos}^{}\bigg[\dfrac{\pi}{5} \bigg]}$

Now,

$\rm :\longmapsto\: {cos}^{3}\bigg[\dfrac{\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{2\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{3\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{4\pi}{5} \bigg]$

can be rewritten as

$\rm = \: {cos}^{3}\bigg[\dfrac{\pi}{5} \bigg] + {cos}^{3}\bigg[\dfrac{2\pi}{5} \bigg] - {cos}^{3}\bigg[\dfrac{2\pi}{5} \bigg] - {cos}^{3}\bigg[\dfrac{\pi}{5} \bigg]$

$\rm \:  =  \:0$

Hence,

• Option (a) is correct

Additional Information :-

$\boxed{ \rm \:sin2x = 2sinxcosx}$

$\boxed{ \rm \:cos2x = 1 - {2sin}^{2} x = {2cos}^{2}x - 1 = {cos}^{2}x - {sin}^{2}x}$

$\boxed{ \rm \:tan2x = \frac{2tanx}{1 - {tan}^{2}x}}$

$\boxed{ \rm \:sin(x + y) = sinxcosy + sinycosx}$

$\boxed{ \rm \:sin(x - y) = sinxcosy - sinycosx}$

$\boxed{ \rm \:cos(x - y) = cosxcosy + sinysinx}$

$\boxed{ \rm \:cos(x + y) = cosxcosy - sinysinx}$