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.





11 and 12 Grade Math

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