promotion image of download ymail app
Promoted

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

I need more help!

6 Answers

Relevance
  • 1 decade ago
    Favorite 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.

    • Commenter avatarLogin to reply the answers
  • Dr. J.
    Lv 6
    1 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.

    • Commenter avatarLogin to reply the answers
  • 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 !!!

    • Commenter avatarLogin to reply the answers
  • 1 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.

    • Commenter avatarLogin to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • Lauren
    Lv 4
    4 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

    • Commenter avatarLogin to reply the answers
  • 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
    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.