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