3 years ago

CosA.cos2A.cos4A=sin8A/8sinA

Mathematics

1 answer:

alexgriva [62]3 years ago

3 0

I assume you're supposed to establish the identity,

cos(A) cos(2A) cos(4A) = 1/8 sin(8A) / sin(A)

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Then you have

sin(8A) = 2 sin(4A) cos(4A)

sin(8A) = 4 sin(2A) cos(2A) cos(4A)

sin(8A) = 8 sin(A) cos(A) cos(2A) cos(4A)

==> sin(8A)/(8 sin(A)) = cos(A) cos(2A) cos(4A)

as required.

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CosA.cos2A.cos4A=sin8A/8sinA​

CosA.cos2A.cos4A=sin8A/8sinA