Identities involving sines and cosines of multiples or submultiples of the angles involved.
To prove the identities involving sines and cosines we use the following algorithm.
Step I: Convert the sum of first two terms as product by using one of the following formulae:
sin C + sin D = 2 sin \(\frac{C + D}{2}\) cos \(\frac{C - D}{2}\)
sin C - sin D = 2 cos \(\frac{C + D}{2}\) sin \(\frac{C - D}{2}\)
cos C + cos D = 2 cos \(\frac{C + D}{2}\) cos \(\frac{C - D}{2}\)
cos C - cos D = - 2 sin \(\frac{C + D}{2}\) sin \(\frac{C - D}{2}\)
Step II: In the product obtain in step II replace the sum of two angles in terms of the third by using the given relation.
Step III: Expand the third term
by using one of the following formulas:
sin 2θ = 2 sin θ cos θ,
cos 2θ = 2 cos\(^{2}\) θ - 1
cos 2θ = 1 - 2 sin\(^{2}\) θ etc.
Step IV: Take the common factor outside.
Step V: Express the trigonometric ratio of the single angle in terms of the remaining angles.
Step VI: Use one of the formulas given in step I to convert the sum into product.
1. If A + B + C = π prove that, sin 2A + sin 2B +
sin 2C = 4 sin A sin B sin C.
Solution:
L.H.S. = (sin 2A + sin 2B) + sin 2C
= 2 sin \(\frac{2A + 2B}{2}\) cos \(\frac{2A - 2B}{2}\)+ sin 2C
= 2 sin (A + B) cos (A - B) + sin 2C
= 2 sin (π - C) cos (A - B) + sin
2C, [Since, A + B + C = π ⇒ A
+ B = π - C]
= 2 sin C cos (A - B) + 2 sin C cos C, [Since sin (π
- C) = sin C]
= 2 sin C [cos (A - B) + cos C], taking common 2 sin C
= 2 sin C [cos (A - B) + cos
{π - (A + B)}], [Since A + B + C = π ⇒ C
= π - (A + B)]
= 2 sin C [cos (A - B) - cos (A + B)], [Since cos {π - (A + B)} = - cos (A + B)]
= 2 sin C [2 sin A sin B], [Since
cos (A - B) - cos (A + B) = 2 sin A sin B]
= 4 sin A sin B sin C. Proved.
2. If A + B + C = π prove that, cos 2A + cos 2B - cos 2C = 1- 4 sin A sin B cos C.
Solution:
L.H.S. = cos 2A + cos 2B - cos 2C
= (cos 2A + cos 2B) - cos 2C
= 2 cos \(\frac{2A + 2B}{2}\) cos \(\frac{2A - 2B}{2}\) - cos 2C
= 2 cos (A + B) cos (A- B) - cos 2C
= 2 cos (π - C) cos (A- B) - cos
2C, [Since we know A + B + C = π ⇒A +
B = π – C]
= - 2 cos C cos (A - B) – (2 cos\(^{2}\) C - 1), [Since cos (π - C) = - cos C]
= - 2 cos C cos (A - B) -
2 cos\(^{2}\) C + 1
= - 2 cos C [cos (A - B) + cos C] + 1
= -2 cos C [cos (A - B) - cos
(A + B)] + 1, [Since cos C = - cos (A + B)]
= -2 cos C [2 sin A sin B] + 1, [Since cos (A - B) - cos (A + B) = 2 sin A sin B]
= 1 - 4 sin A sin B cos C. Proved.
● Conditional Trigonometric Identities
11 and 12 Grade Math
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