EXPRESSING SUM OR DIFFERENCE OF TRIGONOMETRIC FUNCTIONS AS PRODUCT

Some applications of trigonometric functions demand that a sum or difference of trigonometric functions be written as a product of trigonometric functions.

The identities given below will be useful to write the sum and difference of trigonometric functions as product. 

Question 1 :

Express each of the following as a product

(i) sin 75° − sin 35°

Solution :

  =  sin 75° − sin 35°

It exactly matches the formula

sin C - sin D  =  2 cos [(C + D)/2] sin [(C - D)/2]

=  2 cos [(75 + 35)/2] sin [(75 - 35)/2]

=  2 cos (110/2) sin (40/2)

=  2 cos 55° sin 20°

(ii) cos 65° + cos 15°

Solution :

  =  cos 65° + cos 15°

It exactly matches the formula

cos C + cos D  =  2 cos [(C + D)/2] cos [(C - D)/2]

=  2 cos [(65 + 15)/2] cos [(65 - 15)/2]

=  2 cos (80/2) cos (50/2)

=  2 cos 40° cos 25°

(iii) sin 50° + sin 40°

Solution :

  =  sin 50° + sin 40°

It exactly matches the formula

sin C + sin D  =  2 sin [(C + D)/2] cos [(C - D)/2]

=  2 sin [(50 + 40)/2] cos [(50 - 40)/2]

=  2 cos (90/2) cos (10/2)

=  2 cos 45° cos 5°

(iv) cos 35° − cos 75°

Solution :

  =  cos 35° − cos 75°

It exactly matches the formula

cos C - cos D  =  -2 sin [(C + D)/2] sin [(C - D)/2]

=  -2 sin [(35 + 75)/2] cos [(35 - 75)/2]

=  -2 cos (110/2) cos (-40/2)

=  -2 cos 55° cos (-20°)

=  -2 cos 55° cos 20°

cos (-θ)  =  cos θ

Question 2 :

Show that :

sin 12° sin 48° sin 54°  =  1/8

Solution :

  =  sin 12° sin 48° sin 54° 

  =   sin (90 - 36) sin 12° sin 48°

  =   sin (90 - 36) [(-2/-2)(sin 12° sin 48°)]

  =   sin (90 - 36) [(1/-2)(-2sin 12° sin 48°)]

  =   sin (90 - 36) [(1/-2) (cos (12 + 48) - cos (12 - 48))]

  =   cos 36 [(1/-2) (cos 60 - cos 36]

  =  (-1/2) cos 36 [(1/2) - cos 36]

  =  (-1/2) (√5 + 1)/4 [(1/2) -  (√5 + 1)/4] 

  =  (-1/2) (√5 + 1)/4 [(2 - √5 - 1)/4]

  =  (-1/2) (1 + √5)/4 [(1 - √5)/4]

  =  (-1/2) (1 - 5)/16 

  =  (-1/2) (-1/4)

 =  1/8

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