Question

Prove sinA*cos2A=1/4sin4A*secA
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Answer 1

The prove is below

Step-by-step explanation:

From trigonometric identities:

`cos2A=cos(A+A)=cos^2A-sin^2A\\sin2A=sin(A+A)=2sinAcosA\\sin4A=sin(2A+2A)=2sin2Acos2A=2(2sinAcosA)(cos^2A-sin^2A)=4sinAcos^3A-4cosAsin^3A\\secA=(1)/(cosA)`

Therefore:

`(1)/(4)sin4AsecA = (1)/(4)( 4sinAcos^3A-4cosAsin^3A)(1)/(cosA)\\= sinAcos^3A-cosAsin^3A((1)/(cosA) )\\=sinAcos^2-sin^3A\\=sinA(cos^2A-sin^2A)\\But\ cos^2A-sin^2A=cos2A\\Therefore:sinA(cos^2A-sin^2A)=sinAcos2A\\(1)/(4)sin^4Acos2A=sinAcos2A`