How to find the value of sin4x, cos4x, cot4x?

We have: sin(4x)

=sin(2x+2x)

Let’s apply the angle sum identity for 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

cot4x

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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