1.

Prove that: sin2 2π/5 – sin2 π/3 = (√5 – 1)/8

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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