# Express the Product as a Sum or Difference

We will how to express the product as a sum or difference.

1. Convert the product into sum or differences: 2 sin 5x cos 3x

Solution:

2 sin 5x cos 3x = sin (5x + 3x) + sin (5x -3x), [Since 2 sin A cos B = sin (A + B) + sin (A - B)]

= sin 8x + sin 2x

2. Express sin (3∅)/2 ∙ cos (5∅)/2 as sum or difference.

Solution:

sin (3∅)/2  cos (5∅)/2

= 1/2 ∙ 2sin (3∅)/2 cos (5∅)/2

= 1/2 [sin ((3∅)/2 + (5∅)/2) - sin ((5∅)/2 - (3∅)/2)]

= 1/2 (sin 4∅ - sin ∅)

3. Convert 2 cos 5α sin 3α into sum or differences.

Solution:

2 cos 5α sin 3α = sin (5α + 3α) - sin (5α -3α), [Since 2 cos A sin B = sin (A + B) - sin (A - B)]

= sin 8α - sin 2α

4. Express the product as a sum or difference: 4 sin 20° sin 35°

Solution:

4sin 20° sin 35° = 2 ∙ 2 sin20° sin 35°

= 2 [cos (35°- 20°) - cos (35° + 20°)]

= 2 (cos 15° - cos 55°).

5. Convert  cos 9β cos 4β into sum or differences.

Solution:

cos 9β cos 4β = ½ ∙ 2 cos 9β cos 4β

= ½ [cos (9β + 4β) + cos (9β - 4β)], [Since 2 cos A cos B = cos (A + B) + cos (A - B)]

= ½ (cos 13β + cos 5β)

6. Prove that, tan (60° - ∅) tan (60° + ∅) = (2 cos 2∅ + 1)/(2 cos 2∅ - 1)

Solution:

L.H.S. = tan (60° - ∅) tan (60° + ∅)

= (2 sin (60° - ∅) sin (60° + ∅))/(2cos (60° - ∅) cos (60° + ∅)

= cos [(60° + ∅) - (60° - ∅)] - cos [(60° + ∅)+ (60° - ∅) ]/(cos[(60° + ∅ )+ (60° - ∅) ] + cos [(60° + ∅) - (60° - ∅) ] )

= (cos 2∅ - cos 120°)/(cos 120° + cos 2∅)

= (cos 2∅ - (-1/2))/(-1/2 + cos 2∅), [Since cos 120° = -1/2]

= (cos 2∅ + 1/2)/(cos 2∅ - 1/2)

= (2 cos 2∅ + 1)/(2 cos 2∅ - 1)   proved

7. Convert the product into sum or differences: 3 sin 13β sin 3β

Solution:

3 sin 13β sin 3β = 3/2 ∙ 2 sin 13β sin 3β

= 3/2 [cos (13β - 3β) - cos (13β + 3β)], [Since 2 sin A sin B = cos (A - B) - cos (A + B)]

= 3/2 (cos 10β - cos 16β)

8. Show that, 4 sin A sin B sin C = sin (A + B - C) + sin (B + C - A) + sin (C+ A - B) - sin (A + B + C)

Solution:

L.H.S. = 4 sin A sin B sin C

= 2 sin A (2 sin B sin C)

= 2 sin A {cos (B - C) - cos (B + C)}

= 2 sin A ∙ cos (B - C) - 2 sin A cos (B + C)

= sin (A + B - C) + sin (A - B + C) - [sin (A + B + C) - sin (B + C -A)]

= sin (A + B - C) + sin (B + C - A) + sin (A + C - B) - sin (A + B + C) = R.H.S.

Proved