Conditional Identities
You have studied identities like (x +y)² = x² +y² +2xy, which hold for all
values (of x and y). Conditional identity is an identity which holds if
variables satisfy a given condition.
Method of solving conditional identities
- If the identity involves sines and cosines of angles, then
sums/differences should be converted into products using A-B formulae and
then simplification should be done, by using C-D formulae or other relevant formulae.
- If the identity involves squares of sines or cosines, first the squares
should be changed into cosines of double angles by using the formulae
cos²A= (1 +cos2A)/2 and sin²A = (1 -cos2A)/2
- If tangents or cotangents are involved, express the sum of two angles in
terms of third angle, (from given relation) and then take tangents of both
sides and expand, and simplify.
Illustrative Examples
Example
If A, B, C are angles of a triangle, prove that
sin 2A +sin 2C = 4 sin A sin B sin C
Solution
Since A, B, C are angles of a triangle, we have A +B +C =
= 180°.
L.H.S. = (sin 2A +sin 2B) +sin 2C
= 2 sin [(2A +2B)/2] cos [(2A -2B)/2] + sin 2C
= 2 sin (A +B) cos (A -B) + 2 sin C cos C
= 2 sin ( -C) cos (A -B) + 2 sin C cos
( -(A +B))
= 2 sin C cos (A -B) -2 sin C cos (A +B)
= 2 sin C [cos (A -B) -cos (A +B)]
= 2 sin C (2 sin A sin B) = 4 sin A sin B sin C = R.H.S.
Example
If A +B +C = , prove that tan A +tan
B + tan C = tan A tan B tan C
Solution
A +B +C = => A +B =
-C
=> tan (A +B) = tan ( - C)
=> (tan A +tan B)/(1 -tan A tan B) = -tan C
=> tan A +tan B = -tan C +tan A tan B tan C
=> tan A +tan B +tan C = tan A tan B tan C.
Exercise
- If A +B +C = , prove that
(i) sin 2A +sin 2B -sin 2C = 4 cos A cos B sin C
(ii) sin 2A -sin 2B +sin 2C = 4 cos A sin B cos C
(iii) cos 2A +cos 2B +cos 2C = -1 -4 cos A cos B cos C
(iv) cos 2A -cos 2B +cos 2C = -1 -4 sin A cos B sin C.
- If A, B, C are angles of a triangle, prove that
(i) cos²A + cos²B +cos²C = 1 -2 cos A cos B cos C
(ii) sin²A -sin²B + sin²C = 2 sin A cos B sin C
- If A +B +C = , prove that
(i) tan 2A +tan 2B +tan 2C = tan 2A tan 2B tan 2C
(ii) tan A/2 tan B/2 + tan B/2 tan C/2 +tan C/2 tan A/2 = 1
- If A +B +C = , prove that
sin A/2 +sin B/2 +sin C/2 = 1 + 4 sin (
-A)/4 sin ( -B)/4 sin (
-C)/4
- If A +B +C = 90°, prove that
(i) sin 2A +sin 2B +sin 2C = 4 cos A cos B cos C
(ii) cos²A + cos²B +cos²C = 2 +2 sin A sin B sin C
(iii) tan A tan B +tan B tan C +tan C tan A = 1.