sin ^2 54 - sin^2 72= sin^2 18 -sin ^2 36
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Prove that sin²54° -sin²72°= sin²18 °- sin²36°
sin²54° -sin²72°= sin²18 °- sin²36°
To Prove:
sin²54° -sin²72° = sin²18° - sin²36°
Solution:
Consider LHS,
sin²54° -sin²72°
= sin²(90°-36°) - sin²(90°-18°)
= cos²36° - cos²18° (∵sin²A+cos²A=1)
= cos²36° - 1 + sin²18° ------------------------(1)
Consider RHS,
sin²18° -sin²36°
= sin²18° -1 + cos²36° (∵sin²A+cos²A=1)
= cos²36° - 1 + sin²18° ------------------------(2)
From equation (1) and(2)
LHS= RHS
sin²54° -sin²72° = sin²18° - sin²36°
Hence proved
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