How do you prove #[(sin 3x) / sin x ] – [(cos3x) / cos x)] = 2#?

3 Answers

#(sin 3x)/sin x-(cos 3x)/cos x=2" "#TRUE, it is an identity

Explanation:

#(sin 3x)/sin x-(cos 3x)/cos x=2#

#(sin (2x+x))/sin x-(cos (2x+x))/cos x=2#

#(sin 2x*cos x +cos 2x *sin x)/sin x-(cos 2x*cos x- sin 2x*sin x)/cos x=2#

#(sin 2x*cos x)/sin x +(cos 2x *sin x)/sin x-(cos 2x*cos x)/cos x+( sin 2x*sin x)/cos x=2#

#(sin 2x*cos x)/sin x +(cos 2x *cancelsin x)/cancelsin x-(cos 2x*cancelcos x)/cancelcos x+( sin 2x*sin x)/cos x=2#

#(sin 2x*cos x)/sin x +cos 2x -cos 2x+( sin 2x*sin x)/cos x=2#

#(sin 2x*cos x)/sin x +( sin 2x*sin x)/cos x=2#

#(2*sin x*cos x*cos x)/sin x +( 2*sin x*cos x*sin x)/cos x=2#

#(2*cancelsin x*cos x*cos x)/cancelsin x +( 2*sin x*cancelcos x*sin x)/cancelcos x=2#

#(2*cos x*cos x) +( 2*sin x*sin x)=2#

#2*cos^2 x + 2*sin^2 x=2#

#2*(cos^2 x + sin^2 x)=2#

#2*(1)=2#

#2=2" "#it is an IDENTITY

God bless....I hope the explanation is useful.

Jul 1, 2016

#L.H.S.= (sin3xcosx-cos3xsinx)/(sinxcosx)#

Recall here that #sinAcosB-cosAsinB=sin(A-B)#, Later we will also use #sin2x=2sinxcosx#.

#:. L.H.S.=sin(3x-x)/(sinxcosx)=(sin2x)/(sinxcosx)=(2sinxcosx)/(sinxcosx)=2=R.H.S.#

Yet another way to prove this Identity [as, Mr. Leland Adriano Alejandro has rightly said it!] is to use the Multiple Angle Formula for #cos3x=4cos^3x-3cosx#, and, #sin3x=3sinx-4sin^3x#.

#:.L.H.S.= (3sinx-4sin^3x)/sinx-(4cos^3x-3cosx)/cosx#
#={sinx(3-4sin^2x)}/sinx-{cosx(4cos^2x-3)}/cosx#
#=3-4sin^2x-4cos^2x+3#
#=6-4(sin^2x+cos^2x)=6-4=2#, as proved before!

Hope, you will enjoy the proofs!

Jul 1, 2016

As follows

Explanation:

To prove

#(sin 3x)/sin x-(cos 3x)/cos x=2#

Method-1 (shortest)

#LHS=(sin 3x)/sin x-(cos 3x)/cos x#

#=(sin3xcosx-cos3xsinx)/(sinxcosx#

#=sin(3x-x)/(sinxcosx)#

#=sin(2x)/(sinxcosx)=(2sinxcosx)/(sinxcosx)=2#

Using identity #sin2x=2sinx cosx#
proved

Method -2
Using identities

#sin 3x=(3sin x-4sin^3x) and cos3x=(4cos ^3x-3cosx)#

#LHS=(sin 3x)/sin x-(cos 3x)/cos x#

#=(3sin x-4sin^3x)/sin x-(4cos ^3x-3cosx)/cos x#

#=(3sin x)/sinx-(4sin^3x)/sin x-(4cos ^3x)/cosx+(3cosx)/cos x#

#=3-4sin^2x-4cos^2x+3#

#=6-4(sin^2x+cos^2x)=6-4xx1=2#

[ since #(sin^2x+cos^2x)=1#]

proved