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.

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