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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 A2 sin B2 cos C2

Solution:

We have,

A + B + C = π

⇒ C = π - (A + B)

⇒ C2 = π2  - (A+B2)

Therefore, sin (A+B2) = sin (π2  - C2) = cos  C2

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

= (sin A + sin B) - sin C

= 2 sin (A+B2) cos (AB2) - sin C

= 2 sin (πC2) cos (AB2) - sin C

= 2 sin (π2C2) cos AB2 - sin C

= 2 cos C2 cos AB2 - sin C

= 2 cos C2 cos AB2 - 2 sin C2 cos C2

= 2 cos C2[cos AB2 - sin C2]

= 2 cos C2[cos AB2 - sin (π2A+B2)]

= 2 cos C2[cos (AB2) - cos (A+B2)]

= 2 cos C2[cos  (A2B2) - cos  (A2 + B2)]

= 2 cos C2 [(cos A2  cos B2 +  sin A2  sin B2) - (cos A2  cos B2 +  sin A2  sin B2)]

= 2 cos C2[2 sin A2   sin B2]

= 4 sin A2  sin B2 cos C2  = 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 A2 sin B2 sin C2

Solution:

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

Therefore, A + B + C = π

⇒ C = π - (A + B)

C2 = π2  - (A+B2)

Thus, cos (A+B2) = cos (π2  - C2) = sin C2

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

= (cos A + cos B) + cos C  

= 2 cos (A+B2)  cos (AB2)  + cos C

= 2 cos (π2 - C2) cos (AB2)  + cos C

= 2 sin C2 cos (AB2) + 1 - 2 sin2 C2

= 2 sin C2 cos (AB2) - 2 sin2 C2 + 1

= 2 sin C2[cos (AB2)  - sin C2] + 1

= 2 sin C2[cos (AB2)  - sin (π2A+B2)] + 1

= 2 sin C2[cos (AB2)  - cos (A+B2)] + 1

= 2 sin C2 [2 sin A2 sin B2] + 1

= 4 sin C2 sin A2 sin B2 + 1

= 1 + 4 sin A2 sin B2 sin C2                    Proved.


3. If A +  B + C = π prove that, 
sin A2 +sin B2 + sin C2  = 1 + 4 sin πA4  sin πB4  sin πC4  

Solution:

A + B + C = π          

C2 = π2 - A+B2

Therefore, sin C2 = sin (π2  - A+B2)  = cos A+B2

Now, L. H. S. = sin A2 +sin B2 + sin C2

= 2 sin A+B4 cos AB4 + cos (π2 - C2)

= 2 sin πC4 cos AB4 + cos πC2

= 2 sin πC4 cos AB4 + 1 – 2 sin2 πC4

= 2 sin πC4  cos AB4 - 2 sin2 πC4 + 1

= 2 sin πC4 [cos AB4  - sin πC4] + 1

= 2 sin πC4 [cos AB4  - cos {π2  - πC4}] + 1

= 2 sin πC4 [cos AB4 - cos (π4 + C4)] + 1

= 2 sin πC4 [cos AB4 - cos π+C4] + 1

= 2 sin πC4 [2 sin AB+π+C8  sin π+CA+B8] + 1

= 2 sin πC4 [2 sin A+C+πB8 sin B+C+πA8] + 1

= 2 sin πC4 [2 sin πB+πB8 sin πA+πA8] + 1

= 2 sin πC4 [2 sin πB4  sin πA4] + 1

= 4 sin πC4 sin πB4  sin πA4  + 1

= 1 + 4 sin πA4  sin πB4  sin πC4                    Proved.

 

 

4. If A + B + C = π show that, 
cos A2 + cos B2 + cos C2 =  4 cos A+B4 cos B+C4 cos C+A4

Solution:

A + B + C = π   

C2 = π2 - A+B2
Therefore, cos C2 = cos (π2 - A+B2) = sin A+B2

Now, L. H. S. = cos A2 + cos B2 + cos C2

= (cos A2 + cos B2) + cos C2

= 2 cos A+B4  cos AB4 +  sin A+B2  [Since, cos C2  = sin A+B2

= 2 cos A+B4 cos AB4 + 2 sin A+B4 cos A+B4

= 2 cos  A+B4[cos AB4 + sin A+B4]

= 2 cos A+B4 [cos A+B4 + cos (π2 - A+B4)] 

= 2 cos A+B4 [2 cos AB4+π2A+B42 cos  π2A+B4AB42]

= 2 cos A+B4 [2 cos πB4 cos πA4]

= 4 cos A+B4 cos C+A4  cos B+C4, [Since, π - B = A + B + C - B = A + C; Similarly, π - A = B + C] 

= 4 cos A+B4 cos B+C4 cos C+A4.                     Proved.

 Conditional Trigonometric Identities





11 and 12 Grade Math

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