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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 (A−B2) - sin C
= 2 sin (π−C2) cos (A−B2) - sin C
= 2 sin (π2 - C2) cos A−B2 - sin C
= 2 cos C2 cos A−B2 - sin C
= 2 cos C2 cos A−B2 - 2 sin C2 cos C2
= 2 cos C2[cos A−B2 - sin C2]
= 2 cos C2[cos A−B2 - sin (π2 - A+B2)]
= 2 cos C2[cos (A−B2) - cos (A+B2)]
= 2 cos C2[cos (A2 - B2) - 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 (A−B2) + cos C
= 2 cos (π2 - C2) cos (A−B2) + cos C
= 2 sin C2 cos (A−B2) + 1 - 2 sin2 C2
= 2 sin C2 cos (A−B2) - 2 sin2 C2 + 1
= 2 sin C2[cos (A−B2) - sin C2] + 1
= 2 sin C2[cos (A−B2) - sin (π2 - A+B2)] + 1
= 2 sin C2[cos (A−B2) - 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 π−A4 sin π−B4 sin π−C4
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 A−B4 + cos (π2 - C2)
= 2 sin π−C4 cos A−B4 + cos π−C2
= 2 sin π−C4 cos A−B4 + 1 – 2 sin2 π−C4
= 2 sin π−C4 cos A−B4 - 2 sin2 π−C4 + 1
= 2 sin π−C4 [cos A−B4 - sin π−C4] + 1
= 2 sin π−C4 [cos A−B4 - cos {π2 - π−C4}] + 1
= 2 sin π−C4 [cos A−B4 - cos (π4 + C4)] + 1
= 2 sin π−C4 [cos A−B4 - cos π+C4] + 1
= 2 sin π−C4 [2 sin A−B+π+C8 sin π+C−A+B8] + 1
= 2 sin π−C4 [2 sin A+C+π−B8 sin B+C+π−A8] + 1
= 2 sin π−C4 [2 sin π−B+π−B8 sin π−A+π−A8] + 1
= 2 sin π−C4 [2 sin π−B4 sin π−A4] + 1
= 4 sin π−C4 sin π−B4 sin π−A4 + 1
= 1 + 4 sin π−A4 sin π−B4 sin π−C4 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 A−B4 + sin A+B2 [Since, cos C2 = sin A+B2]
= 2 cos A+B4 cos A−B4 + 2 sin A+B4 cos A+B4
= 2 cos A+B4[cos A−B4 + sin A+B4]
= 2 cos A+B4 [cos A+B4 + cos (π2 - A+B4)]
= 2 cos A+B4 [2 cos A−B4+π2−A+B42 cos π2−A+B4−A−B42]
= 2 cos A+B4 [2 cos π−B4 cos π−A4]
= 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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