Question #c2ea6
1 Answer
Future reference:-(see explanation first)
From,
#sinx=(1-sin^2x)/(1+sin^2x)#
#=>sin^2x=(1-sin^2x)^2/(1+sin^2x)^2#
#=>1-cos^2x=(1-sin^2x)^2/(1+sin^2x)^2#
#=>cos^2x=((1+sin^2x)^2-(1-sin^2x)^2)/(1+sin^2x)^2#
#=>cosx=(2sinx)/(1+sin^2x)#
Explanation:
#sinx+sin^2x+sin^3x=1#
#=>sinx+sin^3x=1-sin^2x" "...(2)#
#=>sinx(1+sin^2x)=cos^2x" "...(3)#
#=>sin^2x(1+sin^2x)^2=(cos^2x)^2#
#=>(1-cos^2x)(2-cos^2x)^2=cos^4x#
#=>(1-cos^2x)(4-4cos^2x+cos^4x)=cos^4x#
#=>4-4cos^2x+cos^4x-4cos^2x+4cos^4x-cos^6x=cos^4x#
#=>4-4cos^2x+cancel(cos^4x)-4cos^2x+4cos^4x-cos^6x=cancel(cos^4x#
#=>4-4cos^2x-4cos^2x+4cos^4x-cos^6x=0#
#=>4(1-2cos^2x+cos^4x)=cos^6x#
#=>4(1-cos^2x)^2=cos^6x#
#=>2(1-cos^2x)=cos^3x#
#=>cos^3x+cos^2x=2-cos^2x#
#=>cos^3x+cos^2x+cosx=2-cos^2x+cosx#
#=>cos^3x+cos^2x+cosx=1+sin^2x+(2sinx)/(1+sin^2x)#
#=>cos^3x+cos^2x+cosx=((1+sin^2x)^2+2sinx)/(1+sin^2x)#
#=>cos^3x+cos^2x+cosx=#
