Sum and Difference Identities

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Math Analysis - Trigonometry - Sum and Difference Formulas - Sine - Cosine

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1 of 3 videos by AJ Speller

Key Questions

  • There are 2 identities you must know to simplify your expressions and solve equations

    • sum identities :
      #(a+b)^2= a^2+b^2+2*a*b#
    • difference identities:
      #(a-b)^2= a^2+b^2-2*a*b#

    Proof:
    #(a+b)^2= (a+b)*(a+b)# then you expand
    #(a+b)^2=a^2+b^2+a*b+b*a# and then #a*b=b*a#

    #(a-b)^2= (a-b)*(a-b)# then you expand
    #(a-b)^2=a^2+b^2-a*b-b*a#

    Example:
    #(x+3)^2= x^2+9+6x#
    #(x-3)^2= x^2+9-6x#

    Bonus:
    #(a+b)(a-b)= a^2-b^2#
    proof:
    #(a+b)(a-b)=a^2-a*b+b*a-b^2=a^2-b^2#

  • Here is an example:
    sin(75) = sin(30 + 45) = sin(30) cos(45) + cos(30) sin(45)
    = (1/2) ( #sqrt2#/2) + ( #sqrt3#/2)( #sqrt2#/2)
    = ( #sqrt2# + #sqrt6#)/4

    cos(75) = cos(30 + 45) = cos(30) cos(45) - sin(30) sin(45)
    = ( #sqrt3#/2)( #sqrt2#/2) - (1/2) ( #sqrt2#/2)
    = ( #sqrt6# - #sqrt2#)/4

    sin(15) = sin(45 - 30) = sin(45) cos(30) - cos(45) sin(30)
    = ( #sqrt2#/2) ( #sqrt3#/2) - ( #sqrt2#/2)(1/2)
    = ( #sqrt6# - #sqrt2#)/4

    cos(15) = cos(45 - 30) = cos(45) cos(30) + sin(45) sin(30)
    =( #sqrt2#/2) ( #sqrt3#/2) + ( #sqrt2#/2) (1/2)
    =( #sqrt6# + #sqrt2#)/4

  • Main Sum and Differences Trigonometric Identities

    #cos (a - b) = cos a*cos b + sin a*sin b#
    #cos (a + b) = cos a*cos b - sin a*sin b#
    #sin (a - b) = sin a*cos b - sin b*cos a#
    #sin (a + b) = sin a*cos b + sin b*cos a#
    #tan (a - b) = (tan a - tan b)/(1 + tan a*tan b)#
    #tan (a + b) = (tan a + tan b)/(1 -tan a*tan b)#

    Application of Sum and Differences Trigonometric Identities

    Example 1: Find #sin 2a#.

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

    Example 2: Find #cos 2a#.

    #cos 2a#
    #= cos (a + a)#
    #= cos a*cos a - sin a*sin a#
    #= cos^2 a - sin^2 a#

    Example 3: Find #cos ((13pi)/12)#.

    #cos ((13pi)/12)#
    #= cos (pi/3 + (3pi)/4)#
    #= cos (pi/3)*cos ((3pi)/4) - sin (pi/3)*sin ((3pi)/4)#
    #= -(sqrt2)/4 - (sqrt6)/4#
    #= -[sqrt2 + sqrt6]/4#

  • Here is an example of using a sum identity:

    Find #sin15^@#.

    If we can find (think of) two angles #A# and #B# whose sum or whose difference is 15, and whose sine and cosine we know.

    #sin(A-B)=sinAcosB-cosAsinB#

    We might notice that #75-60=15#
    so #sin15^@=sin(75^@-60^@)=sin75^@cos60^@-cos75^@sin60^@#

    BUT we don't know sine and cosine of #75^@#. So this won't get us the answer. (I included it because when solving problems we DO sometimes think of approaches that won't work. And that's OK.)

    #45-30=15# and I do know the trig functions for #45^@# and #30^@#

    #sin15^@=sin(45^@-30^@)=sin45^@cos30^@-cos45^@sin30^@#

    #=(sqrt2/2)(sqrt3/2)-(sqrt2/2)(1/2)#

    #=(sqrt6 - sqrt 2)/4#

    There are other way of writing the answer.

    Note 1
    We could use the same two angles and the identity for #cos(A-B)# to find #cos 15^@#

    Note 2
    Instead of #45-30=15# we could have used #60-45=15#

    Note 3
    Now that we have #sin 15^@# we could use #60+15=75# and #sin(A+B)# to find #sin75^@#. Although if the question had been to find #sin75^@, I'd probably use #30^@# and #45^@#

Questions

  • Double-check the answer
    Cem Sentin answered · 1 month ago