Identities Involving Sines and Cosines

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.






11 and 12 Grade Math

From Identities Involving Sines and Cosines to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.