# Prove the identity: sin3x/sinx - cos3x/cosx = 2?

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### 6 Answers

- 1 decade agoFavorite Answer
Given: sin(3x) / sin(x) - cos(3x) / cos (x) = 2

To prove this, we manipulate the left-hand side (LHS) of the given equation to reach the right-hand side. That is:

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

= [ sin(3x) cos(x) - cos(3x) sin (x) ] / [ sin(x) cos(x) ].

=sin(3x - x) / [ sin(x) cos(x) ] by applying difference of two angles identity for sine function

=sin(2x) / [ sin(x) cos(x) ] by simplification

= [ 2 sin(x) cos(x) ] / [ sin(x) cos(x) ] by double-angle identity for sine function

= 2 [ sin(x) cos(x) ] / [ sin(x) cos(x) ] by factoring

= 2 by cancellation

= RHS.

This shows that sin(3x) / sin(x) - cos(3x) / cos (x) = 2.

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- Dr. J.Lv 61 decade ago
sin2x = sin(3x-x) = sin3xcosx - sinxcos3x (1)

sin2x = 2sinxcosx (2)

(1) = (2) then

2sinxcosx = sin3xcosx - sinxcos3x

if sinxcosx <>0 then

2 = [sin3xcosx - sinxcos3x] / sinxcosx

2 = sin3x/sinx - cos3x/cosx

Voila.

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- 1 decade ago
hello. ok here's my answer.

sin3x/sinx - cos3x/cosx

= [(sin 3x cos x) - (cos 3x sin x)] / sin x cos x

But we know using the compound angle formulae

sin (3x - x ) = (sin 3x cos x) - (cos 3x sin x)

therefore, sin 2x = (sin 3x cos x) - (cos 3x sin x)

subtituting it back into the equation,

sin3x/sinx - cos3x/cosx

= sin 2x/ six cox

= 2sin 2x / 2sin x cos x

= 2sin 2x / sin 2x

= 2 ( shown )

hope i helped. Good Luck !!!

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- James LLv 51 decade ago
Multiply both sides by sin x cos x:

sin 3x cos x - cos 3x sin x = 2 sin x cos x

2 sin x cos x = sin 2x

and, from the identity sin(A-B) = sin A cos B - cos A sin B, with A=3x and B=x, you get

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

so both sides are equal.

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- LaurenLv 44 years ago
Do you know the formula e(ix) = cos(x) + i*sin(x)? It's very useful for solving these types of problems. You can derive all sorts of relationships with it. In your case e(3ix) = cos(3x) + i*sin(3x) But it's also equal to (e(ix))^3 = (cos(x) + i*sin(x))^3 = cos^3(x) + 3i*cos^2(x)sin(x) - 3cos(x)sin^2(x) - i*sin^3(x) Now compare real and imaginary terms cos(3x) = cos^3(x) - 3cos(x)sin^2(x) sin(3x) = 3cos^2(x)sin(x)-sin^3(x) Now simply substitute: (3cos^2(x)sin(x)-sin^3(x))/sin(x) - (cos^3(x) - 3cos(x)sin^2(x))/cos(x) =3*(sin^2(x)+cos^2(x)) - (sin^2(x)+cos^2(x)) =3-1 = 2

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- 1 decade ago
sin3x/sinx-cos3x/cosx=sin(2x+x)/sinx -cos(2x+x)

sin2x=

Source(s): sorry,I forget that formula at the moment,anyhow,I will tell you asap- Login to reply the answers