Dear Anushka
cos 18° = 4 sin 18° cos 18° (1 - 2sin2 18°)         by the cofunction properties: sin 72° = cos 18°.
            1 = 4 sin 18° (1 - 2sin2 18°)                       Let x = sin 18°, this is known as 
            1 = 4x(1-2x2)                                             substitution, a useful technique in calculus.
8x3-4x+1 = 0                                                         A product is zero only when one of its factors is zero.
8x3-4x+1 = (2x-1)(4x2+2x-1)=0                           (2x-1)=0 implies x= ½=sin 30° > sin 18° ;
                                                                              Since we know sin is increasing on [0°,90°].
            x = (-2 ±  (4 + 4•4•1))/8                       So we must solve the other factor,
                = (-2 ±  20)/8                                     using the quadratic formula.
                = (-2 ±  4 5)/8
                = (-1 ±  5)/4                                       But the sin 18° > 0, so it cannot be negative.
  sin 18°   = ( (5)-1)/4                                         Hence the middle root is the one we want.

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Dear Anushka 
Dont post the same question again.

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….
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