Identities involving sines and cosines of multiples or submultiples of the angles involved.

To prove the identities involving sines and cosines we use the following algorithm.

**Step I:** Convert the sum of first two terms as product by using one of the following formulae:

sin C + sin D = 2 sin \(\frac{C + D}{2}\) cos \(\frac{C - D}{2}\)

sin C - sin D = 2 cos \(\frac{C + D}{2}\) sin \(\frac{C - D}{2}\)

cos C + cos D = 2 cos \(\frac{C + D}{2}\) cos \(\frac{C - D}{2}\)

cos C - cos D = - 2 sin \(\frac{C + D}{2}\) sin \(\frac{C - D}{2}\)

**Step II:** In the product obtain in step II replace the sum of two angles in terms of the third by using the given relation.

**Step III:** Expand the third term
by using one of the following formulas:

sin 2θ = 2 sin θ cos θ,

cos 2θ = 2 cos\(^{2}\) θ - 1

cos 2θ = 1 - 2 sin\(^{2}\) θ etc.

**Step IV:** Take the common factor
outside.

**Step V:** Express the
trigonometric ratio of the single angle in terms of the remaining angles.

**Step VI:** Use one of the formulas
given in step I to convert the sum into product.

Examples on identities involving sines and cosines:

**1.** If A + B + C = π prove that, sin 2A + sin 2B +
sin 2C = 4 sin A sin B sin C. **
**

**Solution: **

L.H.S. = (sin 2A + sin 2B) + sin 2C

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

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

= 2 sin (π - C) cos (A - B) + sin
2C, [Since, A + B + C = π ⇒ A
+ B = π - C]

= 2 sin C cos (A - B) + 2 sin C cos C, [Since sin (π
- C) = sin C]

= 2 sin C [cos (A - B) + cos C], taking common 2 sin C

= 2 sin C [cos (A - B) + cos
{π - (A + B)}], [Since A + B + C = π ⇒ C
= π - (A + B)]

= 2 sin C [cos (A - B) - cos (A + B)], [Since cos {π - (A + B)} = - cos (A + B)]

= 2 sin C [2 sin A sin B], [Since
cos (A - B) - cos (A + B) = 2 sin A sin B]

= 4 sin A sin B sin C. * Proved. *

**
2.** If A + B + C = π prove that, cos 2A + cos 2B - cos 2C = 1- 4 sin A sin B cos C.
**
**

**Solution: **

L.H.S. = cos 2A + cos 2B - cos 2C

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

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

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

= 2 cos (π - C) cos (A- B) - cos
2C, [Since we know A + B + C = π ⇒A +
B = π – C]

= - 2 cos C cos (A - B) – (2 cos\(^{2}\) C - 1), [Since cos (π - C) = - cos C]

= - 2 cos C cos (A - B) -
2 cos\(^{2}\) C + 1

= - 2 cos C [cos (A - B) + cos C] + 1

= -2 cos C [cos (A - B) - cos
(A + B)] + 1, [Since cos C = - cos (A + B)]

= -2 cos C [2 sin A sin B] + 1, [Since cos (A - B) - cos (A + B) = 2 sin A sin B]

= 1 - 4 sin A sin B cos C. * Proved. *

**●**** Conditional Trigonometric Identities**

**Identities Involving Sines and Cosines****Sines and Cosines of Multiples or Submultiples****Identities Involving Squares of Sines and Cosines****Square of Identities Involving Squares of Sines and Cosines****Identities Involving Tangents and Cotangents****Tangents and Cotangents of Multiples or Submultiples**

**11 and 12 Grade Math**

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