PROBLEMS INVOLVING CONDITIONAL IDENTITIES IN TRIGONOMETRY
About "Problems Involving Conditional Identities in Trigonometry"
Problems Involving Conditional Identities in Trigonometry :
Trigonometric identities are true for all admissible values of the angle involved. There are some trigonometric identities which satisfy the given additional conditions. Such identities are called conditional trigonometric identities.
Here we are going to see some example problems to show how to solve conditional trigonometric identities problems.
Problems Involving Conditional Identities in Trigonometry - Examples
Question 1 :
If A + B + C = 2s, then prove that sin(s − A) sin(s − B) + sins sin(s − C) = sin A sin B.
Solution :
A + B + C = 2s
L.H.S :
sin(s − A) sin(s − B) + sins sin(s − C)
sin(s − A) sin(s − B) = (2/2) sin(s − A) sin(s − B)
= (1/2) [cos (s - A - s + B) - cos (s - A + s - B)]
= (1/2) [cos (B - A) - cos (2s - A - B)]
= (1/2) [cos (B - A) - cos (A + B + C - A - B)]
= (1/2) [cos (B - A) - cos C] ------(1)
sins sin(s − C) = (2/2) sins sin(s − C)
= (1/2)(2 sins sin (s - C))
= (1/2)[cos C - cos (2s - C)] ------(2)
(1) + (2)
= (1/2) [cos (B - A)-cos C] + (1/2)[cos C-cos (2s - C)]
= (1/2) [cos (B - A) - cos C + cos C - cos (2s - C)]
= (1/2) [cos (B - A) - cos (A + B + C - C)]
= (1/2) [cos (B - A) - cos (A + B)]
= (1/2) [-2 sin B sin (-A)]
= sin A sin B
Hence proved
Question 2 :
If x + y + z = xyz, then prove that
(2x/1 − x2) + (2y/1 − y2) + (2z/1 − z2) = (2x/1 − x2) (2y/1 − y2) (2z/1 − z2)
Solution :
x + y + z = xyz
Let x = tan A, y = tan B and z = tan C
tan A + tan B + tan C = tan A tan B tan C
tan A + tan B = tan A tan B tan C - tan C
tan A + tan B = tan C (tan A tan B - 1)
tan A + tan B = -tan C (1 - tan A tan B)
(tan A + tan B) / (1 - tan A tan B) = - tan C
tan (A + B) = tan (-C)
A + B = -C
A + B + C = 0
2A + 2B + 2C = 0
tan (2A + 2B) = tan (- 2C)
(tan 2A + tan 2B)/(1 - tan 2A tan 2B) = - tan 2C
(tan 2A + tan 2B) = - tan 2C(1 - tan 2A tan 2B)
tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C-----(1)
tan 2A = 2 tan A / 1 - tan2A = 2x/(1 -x2)
tan 2B = 2 tan B / 1 - tan2B = 2y/(1 -y2)
tan 2C = 2 tan C / 1 - tan2C = 2z/(1 -z2)
By applying those values in (1), we get
2x/(1 -x2) + 2y/(1 -y2) + 2z/(1 - z2) = 2x/(1 -x2) 2y/(1 -y2) 2z/(1 - z2)
Hence proved.
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