Question

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

Answer 1

`(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.

Answer 2

`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!

Answer 3

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