Sines and Cosines of Multiples or Submultiples

We will learn how to solve identities involving sines and cosines of multiples or submultiples of the angles involved.

We use the following ways to solve the identities involving sines and cosines.

(i) Take the first two terms of L.H.S. and express the sum of two sines (or cosines) as product. 

(ii) In the third term of L.H.S. apply the formula of sin 2A (or cos 2A).

(iii) Then use the condition A + B + C = π and take one sine (or cosine) term common. 

(iv) Finally, express the sum or difference of two sines (or cosines) in the brackets as product. 

1. If A + B + C= π   prove that, 

sin A + sin B - sin C = 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) cos \(\frac{C}{2}\)

Solution:

We have,

A + B + C = π

⇒ C = π - (A + B)

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

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

Now, L.H.S. = sin A + sin B - sin C

= (sin A + sin B) - sin C

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

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

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

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

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

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

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

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

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

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

= 2 cos \(\frac{C}{2}\)[2 sin \(\frac{A}{2}\)   sin \(\frac{B}{2}\)]

= 4 sin \(\frac{A}{2}\)  sin \(\frac{B}{2}\) cos \(\frac{C}{2}\)  = R.H.S.                    Proved.

2. If A, B, C be the angles of a triangle, prove that, 

cos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)

Solution:

Since A, B, C are the angles of a triangle,

Therefore, A + B + C = π

⇒ C = π - (A + B)

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

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

Now, L. H. S. = cos A + cos B + cos C

= (cos A + cos B) + cos C  

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

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

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

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

= 2 sin \(\frac{C}{2}\)[cos (\(\frac{A - B}{2}\))  - sin \(\frac{C}{2}\)] + 1

= 2 sin \(\frac{C}{2}\)[cos (\(\frac{A - B}{2}\))  - sin (\(\frac{π}{2}\) -  \(\frac{A + B}{2}\))] + 1

= 2 sin \(\frac{C}{2}\)[cos (\(\frac{A - B}{2}\))  - cos (\(\frac{A + B}{2}\))] + 1

= 2 sin \(\frac{C}{2}\) [2 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\)] + 1

= 4 sin \(\frac{C}{2}\) sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) + 1

= 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)                    Proved.

3. If A +  B + C = π prove that, 
sin \(\frac{A}{2}\) +sin \(\frac{B}{2}\) + sin \(\frac{C}{2}\)  = 1 + 4 sin \(\frac{π - A}{4}\)  sin \(\frac{π - B}{4}\)  sin \(\frac{π - C}{4}\)  

Solution:

A + B + C = π          

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

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

Now, L. H. S. = sin \(\frac{A}{2}\) +sin \(\frac{B}{2}\) + sin \(\frac{C}{2}\)

= 2 sin \(\frac{A + B}{4}\) cos \(\frac{A - B}{4}\) + cos (\(\frac{π}{2}\) - \(\frac{C}{2}\))

= 2 sin \(\frac{π - C}{4}\) cos \(\frac{A - B}{4}\) + cos \(\frac{π - C}{2}\)

= 2 sin \(\frac{π - C}{4}\) cos \(\frac{A - B}{4}\) + 1 – 2 sin\(^{2}\) \(\frac{π - C}{4}\)

= 2 sin \(\frac{π - C}{4}\)  cos \(\frac{A - B}{4}\) - 2 sin\(^{2}\) \(\frac{π - C}{4}\) + 1

= 2 sin \(\frac{π - C}{4}\) [cos \(\frac{A - B}{4}\)  - sin \(\frac{π - C}{4}\)] + 1

= 2 sin \(\frac{π - C}{4}\) [cos \(\frac{A - B}{4}\)  - cos {\(\frac{π}{2}\)  - \(\frac{π - C}{4}\)}] + 1

= 2 sin \(\frac{π - C}{4}\) [cos \(\frac{A - B}{4}\) - cos (\(\frac{π}{4}\) + \(\frac{C}{4}\))] + 1

= 2 sin \(\frac{π - C}{4}\) [cos \(\frac{A - B}{4}\) - cos \(\frac{π + C}{4}\)] + 1

= 2 sin \(\frac{π - C}{4}\) [2 sin \(\frac{A - B + π + C}{8}\)  sin \(\frac{π + C - A + B}{8}\)] + 1

= 2 sin \(\frac{π - C}{4}\) [2 sin \(\frac{A + C + π - B}{8}\) sin \(\frac{B + C + π - A}{8}\)] + 1

= 2 sin \(\frac{π - C}{4}\) [2 sin \(\frac{π - B + π - B}{8}\) sin \(\frac{π - A + π - A}{8}\)] + 1

= 2 sin \(\frac{π - C}{4}\) [2 sin \(\frac{π - B}{4}\)  sin \(\frac{π - A}{4}\)] + 1

= 4 sin \(\frac{π - C}{4}\) sin \(\frac{π - B}{4}\)  sin \(\frac{π - A}{4}\)  + 1

= 1 + 4 sin \(\frac{π - A}{4}\)  sin \(\frac{π - B}{4}\)  sin \(\frac{π - C}{4}\)                    Proved.

 

 

4. If A + B + C = π show that, 
cos \(\frac{A}{2}\) + cos \(\frac{B}{2}\) + cos \(\frac{C}{2}\) =  4 cos \(\frac{A + B}{4}\) cos \(\frac{B + C}{4}\) cos \(\frac{C + A}{4}\)

Solution:

A + B + C = π   

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

Now, L. H. S. = cos \(\frac{A}{2}\) + cos \(\frac{B}{2}\) + cos \(\frac{C}{2}\)

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

= 2 cos \(\frac{A + B}{4}\)  cos \(\frac{A - B}{4}\) +  sin \(\frac{A + B}{2}\)  [Since, cos \(\frac{C}{2}\)  = sin \(\frac{A + B}{2}\)] 

= 2 cos \(\frac{A + B}{4}\) cos \(\frac{A - B}{4}\) + 2 sin \(\frac{A + B}{4}\) cos \(\frac{A + B}{4}\)

= 2 cos  \(\frac{A + B}{4}\)[cos \(\frac{A - B}{4}\) + sin \(\frac{A + B}{4}\)]

= 2 cos \(\frac{A + B}{4}\) [cos \(\frac{A + B}{4}\) + cos (\(\frac{π}{2}\) - \(\frac{A + B}{4}\))] 

= 2 cos \(\frac{A + B}{4}\) [2 cos \(\frac{\frac{A - B}{4} + \frac{π}{2} - \frac{A + B}{4}}{2}\) cos  \(\frac{\frac{π}{2} - \frac{A + B}{4} - \frac{A - B}{4}}{2}\)]

= 2 cos \(\frac{A + B}{4}\) [2 cos \(\frac{π - B}{4}\) cos \(\frac{π - A}{4}\)]

= 4 cos \(\frac{A + B}{4}\) cos \(\frac{C + A}{4}\)  cos \(\frac{B + C}{4}\), [Since, π - B = A + B + C - B = A + C; Similarly, π - A = B + C] 

= 4 cos \(\frac{A + B}{4}\) cos \(\frac{B + C}{4}\) cos \(\frac{C + A}{4}\).                     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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