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Prove that: sin2 2π/5 – sin2 π/3 = (√5 – 1)/8 |
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Answer» Let us consider the LHS sin2 2π/5 – sin2 π/3 = sin2 (π/2 – π/10) – sin2 π/3 As we know, sin (90°– A) = cos A Therefore, sin2 (π/2 – π/10) = cos2 π/10 Sin π/3 = √3/2 Now the above equation becomes, = Cos2 π/10 – (√3/2)2 As we know, cos π/10 = √(10+2√5)/4 Then, the above equation becomes, = [√(10 + 2√5)/4]2 – 3/4 = [10 + 2√5]/16 – 3/4 = [10 + 2√5 – 12]/16 = [2√5 – 2]/16 = [√5 – 1]/8 = RHS Thus proved. |
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