CONDITIONAL IDENTITIES TRIGONOMETRIC EXAMPLES
About "Conditional Identities Trigonometric Examples"
Conditional Identities Trigonometric Examples :
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.
Conditional Identities Trigonometric Examples
Question 1 :
If A + B + C = 180°, prove that
(v) tan A/2 tan B/2 + tan B/2 tan C/2 + tan C/2 tan A/2 = 1
Solution :
L.H.S
= tan A/2 tan B/2 + tan B/2 tan C/2 + tan C/2 tan A/2
A + B + C = 180
A + B = 180 - C
A/2 + B/2 = 90 - C/2
tan (A/2 + B/2) = tan (90 - (C/2))
tan (A/2 + B/2) = cot (C/2)
Using the formula for
tan (A + B) = tan A + tan B / (1 - tan A tan B)
(tan A/2 + tan B/2) / (1 - tan A/2 tan B/2) = 1/tan(C/2)
Take reciprocals on both sides
(1 - tan A/2 tan B/2) / (tan A/2 + tan B/2) = tan (C/2)
1 - tan A/2 tan B/2 = tan (C/2)(tan A/2 + tan B/2)
1 - tan A/2 tan B/2 = tan(A/2)tan(C/2)+tan(B/2)tan(C/2)
1 = tan(A/2)tan(C/2)+tan(B/2)tan(C/2)+tan(A/2)tan(B/2)
tan(A/2)tan(C/2)+tan(B/2)tan(C/2)+tan(A/2)tan(B/2) = 1
Hence proved.
(vi) sinA + sinB + sinC = 4cos A/2 cos B/2 cos C/2
Solution :
= sinA + sinB + sinC
= 2 sin(A + B)/2 cos (A - B)/2 + sin C
= 2 sin(180 - C)/2 cos (A - B)/2 + sin C
= 2 sin C/2 cos (A - B)/2 + 2 sin C/2 cos C/2
= 2 sin C/2 [cos (A - B)/2 + sin C/2]
= 2 sin C/2 [cos (A-B)/2 + sin (180-(A+B))/2]
= 2 sin C/2 [cos (A-B)/2 + sin (90-(A+B)/2)]
= 2 sin C/2 [cos (A-B)/2 + cos (A+B)/2]
= 2 sin C/2 [2 cos A/2 cos B/2]
= 4 cos A/2 cos B/2 sin C/2
(vii) sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4sinAsinB sinC.
Solution :
L.H.S
= sin(B + C − A) + sin(C + A − B) + sin(A + B − C)
A + B + C = 180
|
A + B = 180 - C |
B + C = 180 - A |
C + A = 180 - B |
= sin(180 - A − A) + sin(180 - B − B) + sin(180 - C − C)
= sin(180 - 2A) + sin(180 - 2B) + sin(180 - 2C)
= sin 2A + sin 2B + sin 2C
= 2sin (A + B) cos (A - B) + sin 2C
= 2sin (180-C) cos (A - B) + 2 sin C cos C
= 2sin C cos (A - B) + 2 sin C cos C
= 2 sin C [cos (A - B) + cos C]
= 2 sin C [cos (A - B) + cos (180 - (A + B)]
= 2 sin C [cos (A - B) - cos (A + B)]
= 2 sin C [-2 sin A sin (-B)]
= 4 sin C sin A sin B
= 4 sin A sin B sin C
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