Express the Sum or Difference as a Product

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

1. Convert sin 7α + sin 5α as a product.

Solution:

sin 7α + sin 5α

= 2 sin (7α + 5α)/2 cos (7α - 5α)/2, [Since, sin α + sin β = 2 sin (α + β)/2 cos (α - β)/2]

= 2 sin 6α cos α

2. Express sin 7A + sin 4A as a product.

Solution:

sin 7A + sin 4A

= 2 sin (7A + 4A)/2 cos (7A - 4A)/2

= 2 sin (11A/2) cos (3A)/2

3. Express the sum or difference as a product:  cos ∅ - cos 3∅.

Solution:

cos ∅ - cos 3∅

= 2 sin (∅  + 3∅)/2 sin (3∅ - ∅)/2

= 2 sin 2∅ ∙ sin ∅.

 

4. Express cos 5θ - cos 11θ as a product. 

Solution:

cos 5θ - cos 11θ

= 2 sin (5θ + 11θ)/2 sin (11θ - 5θ), [Since, cos α - cos β = 2 sin (α + β)/2 sin (β - α)/2]

= 2 sin 8θ sin 3θ

 

5. Prove that, sin 55° - cos 55° = √2 sin 10°

Solution:

L.H.S. = sin 55° - cos 55°

           = sin 55° - cos (90° - 35°)

           = sin 55° - sin 35°

           = 2cos (55° + 35°)/2 sin (55° - 35°)/2

           = 2 cos 45° sin 10°

           = 2 ∙ 1/(√2) sin 10°

           = √2 sin 10° = R.H.S.         Proved

6. Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

L.H.S. = sin x + sin 3x + sin 5x + sin 7x

           = (sin 7x + sin x) + (sin 5x + sin 3x)

           = 2 sin (7x + x)/2 cos (7x - x)/2 + 2 sin (5x + 3x)/2 cos (5x - 3x)/2

           = 2 sin 4x cos 3x + 2 sin 4x cos x

           = 2 sin 4x (cos 3x + cos x)

           = 2 sin 4x ∙ 2 cos (3x + x)/2 cos (3x - x)/2

           = 4 sin 4x cos 2x cos x = R.H.S.

 

7. Prove that, sin 20° + sin 140° - cos 10° = 0

Solution:

L.H.S. = sin 20° + sin 140° - cos 10°

           = 2 ∙ sin (140° + 20°)/2 cos (140° - 20°)/2 - cos 10°, [Since sin C + sin D = 2 sin (C + D)/2 cos (C - D)/2]

           = 2 sin 80° ∙ cos 60° - cos 10°

           = 2 ∙ sin (90° - 10°) ∙ 1/2 - cos 10° [Since, cos 60° = 1/2]

           = cos 10° - cos 10°

           = 0 = R.H.S.         Proved

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8. Prove that cos 20° cos 40° cos 80° = 1/8

Solution:

cos 20° cos 40° cos 80°

= ½ cos 40° (2 cos 80° cos 20°)

= ½ cos 40° [cos (80° + 20°) + cos (80° - 20°)]

= ½ cos 40° (cos 100° + cos 60°)

= ½ cos 40° (cos 100° + ½)

= ½ cos 40° cos 100° + ¼ cos 40°

= ¼ (2 cos 40° cos 100°) + ¼ cos 40°

= ¼ [cos (40° + 100°) + cos (40° - 100°)] + ¼ cos 40°

= ¼ [cos 140° + cos (-60°)] + ¼ cos 40°

= ¼ [cos 140° + cos 60°] + ¼ cos 40°

= ¼ [cos 140° + ½] + ¼ cos 40°

= ¼ cos 140° + 1/8 + ¼ cos 40°

= ¼ cos (180° - 40°) + 1/8 + ¼ cos 40°

= - ¼ cos 40° + 1/8 + ¼ cos 40°

= 1/8 = R.H.S.                      Proved

9. Prove that, sin 20° sin 40° sin 60° sin 80°= 3/16

Solution:

L.H.S. = sin 20° ∙ sin 40° ∙ (√3)/2 ∙ sin 80°

           = (√3)/4 ∙ sin 20° (2 sin 40° sin 80°)

           = (√3)/4 ∙ sin 20° [cos (80° - 40°) - cos (80° + 40°)], [Since 2 sin A sin B = cos (A - B) - cos (A + B)]

           = (√3)/4 ∙ sin 20° [cos 40° - cos 120°]

           = (√3)/8 [2 sin 20° cos 40° - 2 sin 20° ∙ (- 1/2)], [Since, cos 120° = cos (180° - 60°) = - cos 60° = -1/2]

           = (√3)/8 [sin (40° + 20°) - sin(40° - 20°) + sin 20°]

           = (√3)/8 [sin 60° - sin 20° + sin 20°]

           = (√3)/8 ∙ (√3)/2 

           = 3/16 = R.H.S.         Proved

10. Prove that, (sin ∅ sin 9∅ + sin 3∅ sin 5∅)/(sin ∅ cos 9∅ + sin 3∅cos 5∅) = tan 6∅

Solution:

 L.H.S. = (sin ∅ sin 9∅+sin 3∅ sin 5∅)/(sin ∅ cos 9∅ +sin 3∅ cos 5∅)

           = (2 sin ∅ sin 9∅ +2 sin 3∅ sin 5∅)/(2 sin ∅ cos 9∅ +2 sin 3∅ cos 5∅)

           = (cos 8∅ - cos 10∅ + cos 2∅ - cos 8∅)/(sin 10∅ - sin 8∅ + sin 8∅ - sin 2∅)  =  (cos 2∅ - cos 10∅)/sin (10 ∅ - sin 2∅)

           = (2 sin 6∅ sin 4∅)/(2 sin 6∅ sin 4∅ ) 

           = tan 6∅    proved

11. Show that 2 cos π/13 cos 9π/13 + cos 3π/13 + cos 5π/13 = 0

Solution:

2 cos π/13 2 cos 9π/13 + cos 3π/13 + cos 5π/13

= 2 cos 9π/13 cos π/13 + cos 3π/13 + cos 5π/13

= cos (9π/13 + π/13) + cos (9π/13 - π/13) + cos 3π/13 + cos 5π/13, [Since, 2 cos X cos Y = cos (X + Y) + cos (X - Y)]

= cos 10π/13 + cos 8π/13 + cos 3π/13 + cos 5π/13

= cos (π - cos 3π/13) + cos (π - cos 5π/13) + cos 3π/13 + cos 5π/13

= - cos 3π/13 - cos 5π/13 + cos 3π/13 + cos 5π/13

= 0

 

12. Express cos A - cos B + cos C - cos (A + B + C) in the product form.

Solution:

(cos A - cos B) + [cos C - cos (A + B + C)]

= 2 sin (A + B)/2 sin (B - A)/2 + 2 sin (C + A + B + C)/2 sin (A + B + C - C)/2

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

= 2 sin (A + B)/2 {2 sin (B - A + A + B + 2C)/4 ∙ cos (A + B + 2C - B + A)/4}

= 4 sin (A + B)/2 sin (B + C)/2 cos (C + A)/2.

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● Converting Product into Sum/Difference and Vice Versa

• Converting Product into Sum or Difference
• Formulae for Converting Product into Sum or Difference
• Converting Sum or Difference into Product
• Formulae for Converting Sum or Difference into Product
• Express the Sum or Difference as a Product
• Express the Product as a Sum or Difference

11 and 12 Grade Math

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