Question

Prove that

${sin}^{2} \frac{\pi}{18} + {sin}^{2} \frac{\pi}{9} + {sin}^{2} \frac{7\pi}{18} + {sin}^{2} \frac{4\pi}{9} = 2$
Complete steps please.​

Answer 1

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

Consider LHS

$\rm \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{\pi}{9} + {sin}^{2} \dfrac{7\pi}{18} + {sin}^{2} \dfrac{4\pi}{9}$

can be further rewritten as

$\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \dfrac{7\pi}{18} + {sin}^{2} \dfrac{8\pi}{18}$

can be further rewritten as

$\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \dfrac{9\pi - 2\pi}{18} + {sin}^{2} \dfrac{9\pi - \pi}{18}$

$\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \bigg(\dfrac{\pi}{2} - \dfrac{2\pi}{18}\bigg) + {sin}^{2}\bigg(\dfrac{\pi}{2} - \dfrac{\pi}{18}\bigg)$

We know,

$\boxed{\rm{  \:{sin}\bigg(\dfrac{\pi}{2} - x\bigg) = cosx\: }}$

So, using this result, we get

$\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {cos}^{2} \dfrac{2\pi}{18} + {cos}^{2}\dfrac{\pi}{18}$

can be further rearrange as

$\rm \: = \: \bigg( {sin}^{2} \dfrac{\pi}{18} + {cos}^{2} \dfrac{\pi}{18}\bigg) + \bigg({sin}^{2} \dfrac{2\pi}{18} + {cos}^{2}\dfrac{2\pi}{18}\bigg)$

$\rm \: =  \:1 + 1$

$\rm \: =  \:2$

Hence,

$\rm\implies \:\boxed{\rm{  \:{sin}^{2} \frac{\pi}{18} + {sin}^{2} \frac{\pi}{9} + {sin}^{2} \frac{7\pi}{18} + {sin}^{2} \frac{4\pi}{9} = 2 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{\rm{  \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\rm{  \:sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }}$

$\boxed{\rm{  \:cosx + cosy = 2cos\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }}$

$\boxed{\rm{  \:cosx - cosy = - 2sin\bigg[\dfrac{x - y}{2} \bigg]sin\bigg[\dfrac{x + y}{2} \bigg] \: }}$

$\boxed{\rm{  \:2sinx \: cosy = sin(x + y) + sin(x - y) \: }}$

$\boxed{\rm{  \:2cosx \: cosy = cos(x + y) + cos(x - y) \: }}$

$\boxed{\rm{  \:2sinx \: siny = cos(x - y) - cos(x + y) \: }}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}} \\ Solution− \\ </p><p></p><p>Consider LHS \\ </p><p></p><p>\begin{gathered}\rm \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{\pi}{9} + {sin}^{2} \dfrac{7\pi}{18} + {sin}^{2} \dfrac{4\pi}{9} \\ \end{gathered}sin218π+sin29π+sin2187π+sin294π \\ </p><p></p><p>can be further rewritten as \\ </p><p></p><p>\begin{gathered}\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \dfrac{7\pi}{18} + {sin}^{2} \dfrac{8\pi}{18} \\ \end{gathered}=sin218π+sin2182π+sin2187π+sin2188π \\ </p><p></p><p>can be further rewritten as \\ </p><p></p><p>\begin{gathered}\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \dfrac{9\pi - 2\pi}{18} + {sin}^{2} \dfrac{9\pi - \pi}{18} \\ \end{gathered}=sin218π+sin2182π+sin2189π−2π+sin2189π−π \\ </p><p></p><p>\begin{gathered}\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {sin}^{2} \bigg(\dfrac{\pi}{2} - \dfrac{2\pi}{18}\bigg) + {sin}^{2}\bigg(\dfrac{\pi}{2} - \dfrac{\pi}{18}\bigg) \\ \end{gathered}=sin218π+sin2182π+sin2(2π−182π)+sin2(2π−18π) \\ </p><p></p><p>We know, \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:{sin}\bigg(\dfrac{\pi}{2} - x\bigg) = cosx\: }} \\ \end{gathered} sin(2π−x)=cosx \\ </p><p></p><p>So, using this result, we get \\ </p><p></p><p>\begin{gathered}\rm \: = \: {sin}^{2} \dfrac{\pi}{18} + {sin}^{2} \dfrac{2\pi}{18} + {cos}^{2} \dfrac{2\pi}{18} + {cos}^{2}\dfrac{\pi}{18} \\ \end{gathered}=sin218π+sin2182π+cos2182π+cos218π \\ </p><p></p><p>can be further rearrange as \\ </p><p></p><p>\begin{gathered}\rm \: = \: \bigg( {sin}^{2} \dfrac{\pi}{18} + {cos}^{2} \dfrac{\pi}{18}\bigg) + \bigg({sin}^{2} \dfrac{2\pi}{18} + {cos}^{2}\dfrac{2\pi}{18}\bigg) \\ \end{gathered}=(sin218π+cos218π)+(sin2182π+cos2182π) \\ </p><p></p><p>\begin{gathered}\rm \: =  \:1 + 1 \\ \end{gathered}= 1+1 \\ </p><p></p><p>\begin{gathered}\rm \: =  \:2 \\ \end{gathered}= 2 \\ </p><p></p><p>Hence, \\ </p><p></p><p>\begin{gathered}\rm\implies \:\boxed{\rm{  \:{sin}^{2} \frac{\pi}{18} + {sin}^{2} \frac{\pi}{9} + {sin}^{2} \frac{7\pi}{18} + {sin}^{2} \frac{4\pi}{9} = 2 \: }} \\ \end{gathered}⟹ sin218π+sin29π+sin2187π+sin294π=2 \\ </p><p></p><p>\rule{190pt}{2pt} \\ </p><p></p><p>Additional Information :- \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }} \\ \end{gathered} sinx+siny=2sin[2x+y]cos[2x−y] \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }} \\ \end{gathered} sinx−siny=2sin[2x−y]cos[2x+y] \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:cosx + cosy = 2cos\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }} \\ \end{gathered} cosx+cosy=2cos[2x−y]cos[2x+y] \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:cosx - cosy = - 2sin\bigg[\dfrac{x - y}{2} \bigg]sin\bigg[\dfrac{x + y}{2} \bigg] \: }} \\ \end{gathered} cosx−cosy=−2sin[2x−y]sin[2x+y] \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:2sinx \: cosy = sin(x + y) + sin(x - y) \: }} \\ \end{gathered} 2sinxcosy=sin(x+y)+sin(x−y) \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:2cosx \: cosy = cos(x + y) + cos(x - y) \: }} \\ \end{gathered} 2cosxcosy=cos(x+y)+cos(x−y) \\ </p><p></p><p>\begin{gathered}\boxed{\rm{  \:2sinx \: siny = cos(x - y) - cos(x + y) \: }} \\ \end{gathered} 2sinxsiny=cos(x−y)−cos(x+y) \\ </p><p></p><p>$