Bagaimana cara mencari nilai sin4x, cos4x, cot4x?
Kita memiliki: sin (4x)
=sin(2x+2x)
Mari kita terapkan identitas jumlah sudut untuk sin(x); sin(α+β)=sin(α)cos(β)+cos(α)sin(β):
=sin(2x)cos(2x)+cos(2x)sin(2x)
=sin(2x)cos(2x)+sin(2x)cos(2x)
=2sin(2x)cos(2x)
cos4x
=cos(2x+2x)
=cos^2 2x-sin^2 2x
=cos^2 2x-(1- cos^2 2x)
=cos^2 2x+cos^2 2x-1
=2cos^2 2x-1
=2cos^2(x+x)-1
=2(cos^2 x-sin^2 x)^2-1
=2(2cos^2 x-1)^2-1
=2[4cos^4 x-4cos^2 x+1)-1
=8(cos x)^4-8(cos x)^2+2-1
=8(cos x)^4-8(cos x)^2+1
ranjang4x
cot4x = 1/tan4x
= 1/tan(x+3x)
= 1/{[tan3x+tanx] / [1-tanx tan3x]}
= [1-tanx tan3x] / [tan3x+tanx]
= [1/tan3x+tanx] – [tan3x*tanx] / [tanx+tan3x]
= [1/tan3x+tanx] – [1/{1/tan3x} + {1/tanx}]
= [1/tan3x+tanx] – [1/cot3x+cotx]
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