Conditional Trigonometric Identities
Subject: Optional Mathematics
Overview
Conditional Trigonometric Identities.
Identities which are true under some given conditions are termed as conditional identities and
in this section, we will deal some trigonometric identities which are bound to the condition of the sum of the angles of a triangle i.e. A + B + C =π
Conditional Trigonometric Identities
examples for Conditional Trigonometric Identities.Identities which are true under some given conditions are termed as conditional identities.
In this section, we will deal some trigonometric identities which are bound to the condition of the sum of the angles of a triangle i.e. A + B + C =π
Properties of supplementary and complementary angles
(i) Since A + B + C = π
Then, A + B = π - C, B + C = π - A and A + C = π - B
Now, sin(A + B) = sin(π - C) = sin C
sin(B + C) = sin( π - A) = sin A
sin(A + C) = sin(π -B) = sin B
Again, cos(A + B) = cos(π - ) = -cos C
cos(B + C) = cos(π - A) = -cos A
cos(A + C) = cos(π - B) = -cos B
Also, tan(A + B) = tan(π- C) = -tan B
tan(B + C) = tan(π- A) = -tan A
tan(A + C) = tan(π - B) = -tan B
(ii) Since A + B + C = π
Then, \(\frac{A}{2}\) + \(\frac{B}{2}\) + \(\frac{C}{2}\) = \(\frac{π}{2}\). So, \(\frac{A + B}{2}\) = \(\frac{π}{2}\) - \(\frac{C}{2}\), \(\frac{B + C}{2}\) = \(\frac{π}{2}\) - \(\frac{A}{2}\) and \(\frac{A + C}{2}\) = \(\frac{π}{2}\) - \(\frac{B}{2}\)
Now, sin(\(\frac{A + B}{2}\)) = sin(\(\frac{π}{2}\) - \(\frac{C}{2}\)) = cos \(\frac{C}{2}\)
sin(\(\frac{B + C}{2}\)) = sin(\(\frac{π}{2}\) - \(\frac{A}{2}\)) = cos \(\frac{A}{2}\)
sin(\(\frac{A + C}{2}\)) = sin(\(\frac{π}{2}\) - \(\frac{B}{2}\)) = cos \(\frac{B}{2}\)
Again, cos(\(\frac{A + B}{2}\)) = cos(\(\frac{π}{2}\) - \(\frac{C}{2}\)) = sin \(\frac{C}{2}\)
cos(\(\frac{A + C}{2}\)) = cos(\(\frac{π}{2}\) - \(\frac{B}{2}\)) = sin \(\frac{B}{2}\)
cos(\(\frac{B + C}{2}\)) = cos(\(\frac{π}{2}\) - \(\frac{A}{2}\)) = sin \(\frac{A}{2}\)
Also, tan(\(\frac{A +B}{2}\)) = tan(\(\frac{π}{2}\) - \(\frac{C}{2}\)) = cot \(\frac{C}{2}\)
tan(\(\frac{A + C}{2}\)) = tan(\(\frac{π}{2}\) - \(\frac{B}{2}\)) = cot \(\frac{B}{2}\)
tan(\(\frac{B + C}{2}\)) = tan(\(\frac{π}{2}\) - \(\frac{B}{2}\)) = cot \(\frac{A}{2}\)
Things to remember
Conditional Trigonometric Identities
- Properties of supplementary and complementary angles
- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.
Videos for Conditional Trigonometric Identities
Solving Trigonometric Equations
Trigonometric Identities
Trigonometry : Proving trig Identities
Questions and Answers
If A + B + C = 180°, prove that:
cos 2A + cos 2B - cos 2C = 1 - 4 sinA sinB cosC
Here,
A + B + C = 180°
or, A + B = 180° - C
cos (A + B) = cos (180° - C) = - cosC
sin (A + B) = sin (180° - C) = sinC
L.H.S.
= cos 2A + cos 2B - cos 2C
= 2 cos\(\frac {2A + 2B}2\) cos\(\frac {2A - 2B}2\) - (2 cos2C - 1)
= 2 cos(A + B) cos (A - B) - 2 cos2C + 1
= 2 (- cosC) cos (A - B) - 2 cos2C+ 1
= 1 - 2 cosC cos(A - B) - 2 cos2C
= 1 - 2 cosC [cos (A - B) + cosC]
= 1 - 2 cosC [cos (A - B) - cos (A + B)]
= 1 - 2 cosC [cosA cosB + sinA sinB - cosA cosB + sinA sinB]
= 1 - 2 cosC× 2 sinA sinB
= 1 - 4 sinA sinB cosC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
cos 2A - cos 2B + cos 2C = 1 - 4 sinA cosB sinC
Here,
A + B + C = 180°
or, A + B = 180° - C
cos (A + B) = cos (180° - C) = - cosC
sin (A + B) = sin (180° - C) = sinC
L.H.S.
= cos 2A - cos 2B + cos 2C
= - 2 sin\(\frac {2A + 2B}2\) sin\(\frac {2A - 2B}2\) + 1 - 2 sin2C
= - 2 sin (A + B) sin (A - B) - 2 sin2C + 1
= - 2 sinC sin (A - B) - 2 sin2C + 1
= - 2 sinC [sin (A - B) - sinC] + 1
= - 2 sinC [sin (A - B) + sin (A + B)] + 1
= - 2 sinC [2 sin\(\frac {A - B + A + B}2\) cos\(\frac {A - B - A - B}2\)] + 1
= - 2 sinC× 2 sinA cos (-B) + 1
= 1 - 4 sinA cosB sinC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
cos 2A + cos 2B + cos 2C + 1 = - 4 cosA cosB cosC
Here,
A + B + C = 180°
or, A + B = 180° - C
cos (A + B) = cos (180° - C) = - cosC
sin (A + B) = sin (180° - C) = sinC
L.H.S.
= cos 2A + cos 2B + cos 2C + 1
= 2 cos\(\frac {2A + 2B}2\) cos\(\frac {2A - 2B}2\) + cos 2C + 1
= 2 cos (A + B) cos (A - B) + 2 cos2C - 1 + 1
= - 2 cosC [cos (A - B) - cosC]
= - 2 cosC [cos (A - B) + cos (A - B)]
= - 2 cosC× 2 cosA cosB
= - 4 cosA cosB cosC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = 180°, prove that:
cos 2A - cos 2B - cos 2C = 4 cosA sinB sinC - 1
Here,
A + B + C = 180°
or, A + B = 180° - C
cos (A + B) = cos (180° - C) = - cosC
sin (A + B) = sin (180° - C) = sinC
L.H.S.
= cos 2A - cos 2B - cos 2C
= - 2 sin\(\frac {2A + 2B}2\) sin\(\frac {2A - 2B}2\) - (1 - 2 sin2C)
= - 2 sin (A + B) sin (A - B) - 1 + 2 sin2C
= - 2 sinC sin (A - B) + 2 sin2C - 1
= - 2 sinC [sin (A - B) - sinC] - 1
= - 2 sinC [sin (A - B) - sin (A + B)] - 1
= - 2 sinC [2 cos\(\frac {A - B + A + B}2\) . sin\(\frac {A - B - A -B}2\)] - 1
= - 2 sinC [2 cosA sin (-B)] - 1
= - 2 sinC× 2 cosA (- sinB) - 1
= 4 cosA sinB sinC - 1
Hence, L.H.S. = R.H.S. Proved
If A, B, C are the angles of a triangle ABC, prove that:
sinA + sinB - sinC = 4 sin\(\frac A2\) sin\(\frac B2\) cos\(\frac C2\)
Here,
A + B + C = \(\pi^c\)
or, A + B = \(\pi^c\) - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {\pi^c}2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac {\pi^c}2\) - \(\frac C2\)) = cos\(\frac C2\)
cos (\(\frac A2\) + \(\frac B2\)) = cos (\(\frac {\pi^c}2\) - \(\frac C2\)) = sin\(\frac C2\)
L.H.S.
= sinA + sinB - sinC
= 2 sin\(\frac {A + B}2\) cos\(\frac {A - B}2\) - sinC
= 2 cos\(\frac C2\) cos\(\frac {A - B}2\) - 2 sin\(\frac C2\) cos\(\frac C2\)
= 2 cos\(\frac C2\) [cos\(\frac {A -B}2\) - sin\(\frac C2\)]
= 2 cos\(\frac C2\) [cos\(\frac {A - B}2\) - cos\(\frac {A + B}2\)] [\(\because\) sin\(\frac C2\) = cos\(\frac {A + B}2\)]
= 2 cos\(\frac C2\) [\(\frac {2 sin\frac A2 - \frac B2 + \frac A2 + \frac B2}{2}\) × \(\frac {sin \frac A2 + \frac B2 - \frac A2 + \frac B2}2\)]
= 2 cos\(\frac C2\)× 2 sin \(\frac {2A}2\)× \(\frac 12\) sin\(\frac {2B}2\)× \(\frac 12\)
= 4 cos\(\frac C2\) sin\(\frac A2\) sin\(\frac B2\)
=4 sin\(\frac A2\) sin\(\frac B2\) cos\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
cosA + cosB + cosC = 1 + 4 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Here,
A + B + C =180°
or, A + B =180° - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {180°}2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac {180°}2\) - \(\frac C2\)) = cos\(\frac C2\)
cos (\(\frac A2\) + \(\frac B2\)) = cos (\(\frac {180°}2\) - \(\frac C2\)) = sin\(\frac C2\)
L.H.S.
= cosA + cosB + cosC
= 2 cos\(\frac {A + B}2\) cos\(\frac {A - B}2\) + 1 - 2 sin2\(\frac C2\)
= 2 sin\(\frac C2\)cos\(\frac {A - B}2\) - 2 sin2\(\frac C2\)+ 1
= 2 sin\(\frac C2\) [cos\(\frac {A - B}2\) - sin\(\frac C2\)] + 1
= 2 sin\(\frac C2\) [cos\(\frac {A - B}2\) - cos\(\frac {A - B}2\)] + 1
= 2 sin\(\frac C2\) [-2 sin\(\frac {\frac A2 - \frac B2 + \frac A2 + \frac B2}2\) sin\(\frac {\frac A2 - \frac B2 - \frac A2 - \frac B2}2\)] + 1
= 2 sin\(\frac C2\) [2 sin\(\frac A2\) sin (-\(\frac B2\))] + 1
=2 sin\(\frac C2\) × 2 sin\(\frac A2\) sin\(\frac B2\) + 1 [\(\because\) sin (-\(\theta\)) = - sin\(\theta\)]
= 1 + 4 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B C = 180°, prove that:
cosB + cosC - cosA = 4 sin\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\) - 1
Here,
A + B + C =180°
or, B+ C =180° - A
or, \(\frac B2\) + \(\frac C2\) = \(\frac {180°}2\) - \(\frac A2\)
sin (\(\frac B2\) + \(\frac C2\)) = sin (\(\frac {180°}2\) - \(\frac A2\)) = cos\(\frac A2\)
cos (\(\frac B2\) + \(\frac C2\)) = cos (\(\frac {180°}2\) - \(\frac A2\)) = sin\(\frac A2\)
L.H.S.
= cosB + cosC - cosA
= 2 cos\(\frac {B + C}2\) cos\(\frac {B - C}2\) - (1 - 2 sin2\(\frac A2\))
= 2 sin\(\frac A2\) cos\(\frac {B - C}2\) - 1 + 2 sin2\(\frac A2\)
= 2 sin\(\frac A2\) cos\(\frac {B - C}2\) + 2 sin2\(\frac A2\) - 1
= 2 sin\(\frac A2\) [cos\(\frac {B - C}2\) + sin\(\frac A2\)] - 1
= 2 sin\(\frac A2\) [cos\(\frac {B - C}2\) + cos\(\frac {B + C}2\)] - 1
= 2 sin\(\frac A2\) [2 cos\(\frac {\frac B2 - \frac C2 + \frac B2 + \frac C2}2\) cos\(\frac {\frac B2 - \frac C2 - \frac B2 - \frac C2}2\)] - 1
= 2 sin\(\frac A2\) [2 cos2\(\frac {2B}2\)× \(\frac 12\) cos -\(\frac {2C}2\) × \(\frac 12\)] - 1
= 2 sin\(\frac A2\) [2 cos2\(\frac B2\) cos\(\frac C2\)] - 1 [\(\because\) cos (-\(\theta\)) = cos\(\theta\)]
=4 sin\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\) - 1
Hence, L.H.S. = R.H.S. Proved
If A, B and C are the vertices of the triangle ABC, prove that:
cot\(\frac A2\) cot\(\frac B2\) cot\(\frac C2\) = cot\(\frac A2\) + cot\(\frac B2\) + cot\(\frac C2\)
Here,
A + B + C = 180°
or, A + B = 180° - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {180°}2\) - \(\frac C2\)
Putting cot on both sides,
cot (\(\frac A2\) + \(\frac B2\)) = cot (90° - \(\frac C2\))
or, \(\frac {cot\frac A2 cot\frac B2 - 1}{cot\frac A2 + cot\frac B2}\) = tan\(\frac C2\)
or, \(\frac {cot\frac A2 cot\frac B2 - 1}{cot\frac A2 + cot\frac B2}\) = \(\frac 1{cot\frac C2} \)
or, cot\(\frac A2\) cot\(\frac B2\)cot\(\frac C2\) - cot\(\frac C2\) = cot\(\frac A2\) + cot\(\frac B2\)
∴ cot\(\frac A2\) cot\(\frac B2\)cot\(\frac C2\) =cot\(\frac A2\) + cot\(\frac B2\) +cot\(\frac C2\)
Hence, L.H.S. = R.H.S. proved
Given that A, B and C are the angles of \(\triangle\) ABC, prove that:
tan\(\frac A2\) tan\(\frac B2\) + tan\(\frac B2\) tan\(\frac C2\) + tan\(\frac C2\) tan\(\frac A2\) = 1
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {\pi}2\) - \(\frac C2\)
Putting tan on both sides,
tan (\(\frac A2\) + \(\frac B2\)) = tan (\(\frac {\pi}2\) - \(\frac C2\))
or, \(\frac {tan\frac A2 + tan\frac B2}{1 - tan\frac A2 tan\frac B2}\) = cot\(\frac C2\)
or,\(\frac {tan\frac A2 + tan\frac B2}{1 - tan\frac A2 tan\frac B2}\) = \(\frac 1{tan\frac C2}\)
or, tan\(\frac A2\) tan\(\frac C2\) + tan\(\frac B2\) tan\(\frac C2\) = 1 - tan\(\frac A2\) tan\(\frac B2\)
∴ tan\(\frac A2\) tan\(\frac C2\) + tan\(\frac B2\) tan\(\frac C2\) +tan\(\frac A2\) tan\(\frac B2\) = 1
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
sinB + sinC - sinA = 4 cos\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Here,
A + B + C =\(\pi\)
or, B+ C =\(\pi\) - A
or, \(\frac B2\) + \(\frac C2\) = \(\frac {\pi}2\) - \(\frac A2\)
sin (\(\frac B2\) + \(\frac C2\)) = sin (\(\frac {\pi}2\) - \(\frac A2\)) = cos\(\frac A2\)
cos (\(\frac B2\) + \(\frac C2\)) = cos (\(\frac {\pi}2\) - \(\frac A2\)) = sin\(\frac A2\)
L.H.S.
=sinB + sinC - sinA
= 2 sin\(\frac {B + C}2\) cos\(\frac {B - C}2\) - 2 sin\(\frac A2\) cos\(\frac A2\)
= 2 cos\(\frac A2\) cos\(\frac {B - C}2\) - 2 sin\(\frac A2\) cos\(\frac A2\)
= 2 cos\(\frac A2\) [cos\(\frac {B - C}2\) - sin\(\frac A2\)]
= 2 cos\(\frac A2\) [cos\(\frac {B - C}2\) - cos\(\frac {B + C}2\)]
= 2 cos\(\frac A2\) [2 sin\(\frac {\frac B2 - \frac C2 + \frac B2 + \frac C2}2\) sin\(\frac {\frac B2 + \frac C2 - \frac B2 + \frac C2}2\)]
= 2 cos\(\frac A2\) [2 sin\(\frac {2B}2\) × \(\frac 12\) sin\(\frac {2C}2\) × \(\frac 12\)]
= 4 cos\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If \(\angle\)A + \(\angle\)B + \(\angle\)C = 180°, prove that:
sinA + sinB + sinC = 4 cos\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\)
Here,
A + B + C =180°
or, A + B =180° - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {180°}2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac {180°}2\) - \(\frac C2\)) = cos\(\frac C2\)
cos (\(\frac A2\) + \(\frac B2\)) = cos (\(\frac {180°}2\) - \(\frac C2\)) = sin\(\frac C2\)
L.H.S.
= sinA + sinB + sinc
= 2 sin\(\frac {A + B}2\) cos\(\frac {A - B}2\) + 2 sin\(\frac C2\) cos\(\frac C2\)
= 2 cos\(\frac C2\) cos\(\frac {A - B}2\) + 2 sin\(\frac C2\) cos\(\frac C2\) [\(\because\) sin\(\frac {A + B}2\) = sin\(\frac C2\)]
= 2 cos\(\frac C2\) [cos\(\frac {A - B}2\) + sin\(\frac C2\)]
= 2 cos\(\frac C2\) [cos\(\frac {A - B}2\) + cos\(\frac {A + B}2\)] [\(\because\) cos\(\frac {A + B}2\) = sin\(\frac C2\)]
= 2 cos\(\frac C2\) [2 cos\(\frac {\frac A2 - \frac B2 + \frac A2 + \frac B2}{2}\) × cos\(\frac {\frac A2 - \frac B2 - \frac A2 -\frac B2}2\)]
= 4 cos\(\frac C2\) cos\(\frac {2A}2\)× \(\frac 12\) cos\(\frac {-2B}2\)× \(\frac 12\)
= 4cos\(\frac C2\) cos\(\frac A2\) cos\(\frac B2\) [\(\because\) cos(- \(\theta\)) = cos\(\theta\)]
=4 cos\(\frac A2\) cos\(\frac B2\)cos\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos (\(\pi\) - C) = - cosC
L.H.S.
=sin 2A + sin 2B + sin 2C
= 2 sin (\(\frac {2A + 2B}2\)) . cos (\(\frac {2A - 2B}2\)) + 2 sinC cosC
= 2 sin (A + B) . cos (A - B) + 2 sinC cosC
= 2 sinC . cos (A - B) + 2 sinC cosC
= 2 sinC [cos (A - B) -cos (A + B)]
= 2 sinC [cos (A - B) - cosC]
= 2 sinC [2 sin(\(\frac {A - B + A + A}2\)) . sin(\(\frac {A + B - (A - B)}2)\)]
= 2 sinC [2 sinA . sin(\(\frac {A + B - A + B}2\))]
= 2 sinC [2 sinA sinB]
= 4 sinA sinB sinC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\)c, prove that:
sin 2A + sin 2B - sin2C = 4 cosA cosB sinC
Here,
A + B + C = \(\pi\)c
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos (\(\pi\) - C) = - cosC
L.H.S.
= sin 2A + sin 2B - sin 2C
= 2 sin\(\frac {2A + 2B}2\) cos\(\frac {2A - 2B}2\) - 2 sinC cosC
= 2 sin (A + B) cos (A - B) - 2 sinC cosC
= 2 sinC . cos (A - B) - 2 sinC cosC
= 2 sinC [cos (A - B) - cosC]
= 2 sinC [cos (A - B) + cos (A + B)]
= 2 sinC [2 cos\(\frac {A - B + A + B}2\) . cos\(\frac {A - B - (A + B)}2\)]
= 2 sinC [2 cos\(\frac {2A}2\) . cos\(\frac {2B}2\)]
= 4 cosA cosB sinC
Hence, L.H.S. = R.H.S. Proved
If \(\alpha\) + \(\beta\) + \(\gamma\) = 180°, prove that:
sin2\(\alpha\) + sin2\(\beta\) + sin2\(\gamma\) = 2 +2 cos\(\alpha\) cos\(\beta\) cos\(\gamma\)
Here,
\(\alpha\) + \(\beta\) + \(\gamma\) = 180°
or, \(\alpha\) + \(\beta\) = 180° - \(\gamma\)
sin (\(\alpha\) + \(\beta\)) = sin (180° - \(\gamma\)) = sin\(\gamma\)
cos (\(\alpha\) + \(\beta\)) = cos (180° - \(\gamma\)) = - cos\(\gamma\)
L.H.S.
=sin2\(\alpha\) + sin2\(\beta\) + sin2\(\gamma\)
= \(\frac {1 - cos 2\alpha}2\) + \(\frac {1 - cos 2\beta}2\) + sin2\(\gamma\)
= \(\frac 12\) - \(\frac 12\) cos 2\(\alpha\) + \(\frac 12\) - \(\frac 12\) cos 2\(\beta\) + sin2\(\gamma\)
= 1 - \(\frac 12\) [cos 2\(\alpha\) + cos 2\(\beta\)] + sin2\(\gamma\)
= 1 - \(\frac 12\) [2 cos\(\frac {2\alpha + 2\beta}2\) . cos\(\frac {2\alpha - 2\beta}2\)] + sin2\(\gamma\)
= 1 - \(\frac 12\)× 2 cos (\(\alpha\) + \(\beta\)) . cos (\(\alpha\) - \(\beta\)) + sin2\(\gamma\)
= 1 + cos\(\gamma\) . cos (\(\alpha\) - \(\beta\)) + 1 - cos2\(\gamma\)
= 2 + cos\(\gamma\) [cos (\(\alpha\) - \(\beta\)) - cos\(\gamma\)]
= 2 + cos\(\gamma\) [cos (\(\alpha\) - \(\beta\)) +cos (\(\alpha\) + \(\beta\))]
= 2 + cos\(\gamma\) [2 cos\(\frac {\alpha - \beta + \alpha + \beta}2\) . cos\(\frac {\alpha - \beta - \alpha - \beta}2\)]
= 2 + cos\(\gamma\)× 2 cos\(\frac {2\alpha}2\) . cos(\(\frac {-2\beta}2\))
= 2 + 2 cos\(\alpha\) . cos\(\beta\) . cos\(\gamma\)
Hence, L.H.S. = R.H.S Proved
If A + B + C = 180° then prove that:
sin2\(\frac A2\) + sin2\(\frac B2\) + sin2\(\frac C2\) = 1 - 2 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Here,
A + B + C = 180°
or, A + B = 180° - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {180°}2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac {180°}2\) - \(\frac C2\)) = cos\(\frac C2\)
cos (\(\frac A2\) + \(\frac B2\)) = cos (\(\frac {180°}2\) - \(\frac C2\)) = sin \(\frac C2\)
L.H.S.
=sin2\(\frac A2\) + sin2\(\frac B2\) + sin2\(\frac C2\)
= \(\frac {1 - cosA}2\) + \(\frac {1 - cosB}2\) + sin2\(\frac C2\)
= \(\frac 12\) - \(\frac 12\) cosA + \(\frac 12\) - \(\frac 12\) cosB + sin2\(\frac C2\)
= 1 - \(\frac 12\) [cosA + cosB] + sin2\(\frac C2\)
= 1 - \(\frac 12\)× 2 cos\(\frac {A + B}2\) . cos\(\frac {A - B}2\) + sin2\(\frac C2\)
= 1 - sin\(\frac C2\) . cos\(\frac {A - B}2\) + sin2\(\frac C2\)
= 1 - sin\(\frac C2\) [cos\(\frac {A - B}2\) - sin\(\frac C2\)]
= 1 - sin\(\frac C2\) [cos\(\frac {A - B}2\) - cos\(\frac {A + B}2\)]
= 1 - sin\(\frac C2\) [2 sin\(\frac {\frac A2 - \frac B2 + \frac A2 + \frac B2}2\) . sin\(\frac {\frac A2 + \frac B2 - \frac A2 + \frac B2}2\)]
= 1 - sin\(\frac C2\)× 2 sin\(\frac {2A}2\)× \(\frac 12\) sin\(\frac {2B}2\)× \(\frac 12\)
=1 - 2 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = 180°, prove that:
cos2\(\frac A2\) + cos2\(\frac B2\) - cos2\(\frac C2\) = 2 cos\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\)
Here,
A + B + C = 180°
or, A + B = 180° - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac {180°}2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac {180°}2\) - \(\frac C2\)) = cos\(\frac C2\)
cos(\(\frac A2\) + \(\frac B2\)) = cos (\(\frac {180°}2\) - \(\frac C2\)) = sin\(\frac C2\)
L.H.S.
=cos2\(\frac A2\) + cos2\(\frac B2\) - cos2\(\frac C2\)
= \(\frac {1 + cosA}2\) + \(\frac {1 + cosB}2\) -cos2\(\frac C2\)
= \(\frac 12\) + \(\frac 12\) cosA + \(\frac 12\) + \(\frac 12\) cosB -cos2\(\frac C2\)
= 1 + \(\frac 12\) (cosA + cosB) - (1 - sin2\(\frac C2\))
= 1 + \(\frac 12\)× 2 cos\(\frac {A + B}2\) cos\(\frac {A - B}2\) - 1 + sin2\(\frac C2\)
= sin\(\frac C2\) . cos\(\frac {A - B}2\) + sin2\(\frac C2\)
= sin\(\frac C2\) (cos\(\frac {A - B}2\) + sin\(\frac C2\))
= sin\(\frac C2\) (cos\(\frac {A - B}2\) + cos\(\frac {A + B}2\))
= sin\(\frac C2\) [2cos\(\frac {\frac A2 - \frac B2 + \frac A2 + \frac B2}2\) × cos\(\frac {\frac A2 - \frac B2 - \frac A2 - \frac B2}2\)]
= sin\(\frac C2\)× 2 . cos\(\frac {2A}2\)× \(\frac 12\) cos (\(\frac {-2B}2\))× \(\frac 12\)
= 2 cos\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\)c, prove that:
cos2A + cos2B - sin2C = - 2 cosA cosB cosC
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos(\(\pi\) - C) = - cosC
L.H.S.
=cos2A + cos2B - sin2C
= \(\frac 12\) [2 cos2A + 2 cos2B] - (1 - cos2C)
= \(\frac 12\) [1 + cos 2A + 1 + cos 2B] + cos2C - 1
= \(\frac 12\) [2 + cos 2A + cos 2B] + cos2C - 1
= \(\frac 12\)× 2 + \(\frac 12\) [cos 2A + cos 2B]+ cos2C - 1
= 1 + \(\frac 12\) [2 cos(\(\frac {2A + 2B}2\)) cos(\(\frac {2A - 2B}2\))]+ cos2C - 1
= \(\frac 12\)× 2 cos 2(\(\frac {A + B}2\)) cos 2(\(\frac {A - B}2\)) + cos2C
= cos (A + B) cos (A - B)+ cos2C
= - cosC cos (A + B) + cos2C
= - cosC [cos (A - B) - cosC]
= - cosC [cos (A - B) + cos (A + B)]
= - cosC 2 cosA cosB
= - 2 cosA cosB cosC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = 180°. prove that:
sin\(\frac A2\) + sin\(\frac B2\) + sin\(\frac C2\) = 1 + 4 sin\(\frac {\pi - A}4\) sin\(\frac {\pi - B}4\) sin\(\frac {\pi - C}4\)
Here,
A + B + C = 180°
or, A + B = 180° - C
or, \(\frac A4\) +\(\frac B4\) = \(\frac {180°}4\) - \(\frac C4\)
sin (\(\frac A4\) +\(\frac B4\)) = sin (\(\frac {180°}4\) - \(\frac C4\))
A + B + C = \(\pi\)
A + C = \(\pi\) - B
B + C = \(\pi\) - A
L.H.S.
=sin\(\frac A2\) + sin\(\frac B2\) + sin\(\frac C2\)
= 1 +sin\(\frac A2\) + sin\(\frac B2\) + sin\(\frac C2\) - 1
= 1 + 2 sin(\(\frac {A + B}4\)) cos\(\frac {A - B}4\) + sin\(\frac C2\) - sin\(\frac {\pi}2\)
= 1 + 2 sin(\(\frac {A + B}4\)) . cos(\(\frac {A - B}4\)) + 2 cos(\(\frac {C + \pi}4\)) . sin (\(\frac {C - \pi}4\))
= 1 + 2 sin(\(\frac {A + B}4\)) [cos\(\frac {A - B}4\) - cos\(\frac {\pi + C}2\)]
=1 + 2 sin(\(\frac {A + B}4\)) [2 sin\(\frac {A - B + \pi + C}8\) sin\(\frac {\pi + C - A + B}8\)]
=1 + 2 sin(\(\frac {A + B}4\)) [2 sin\(\frac {\pi - B + A + C}8\) sin\(\frac {\pi - A + B + C}8\)]
=1 + 2 sin(\(\frac {A + B}4\)) [2 sin\(\frac {\pi - B + \pi - B}8\) sin\(\frac {\pi - A + \pi - A}8\)]
=1 + 2 sin(\(\frac {A + B}4\)) [2 sin\(\frac {2(\pi - B)}8\) sin\(\frac {2(\pi - A)}8\)]
=1 + 2 sin(\(\frac {A + B}4\))× 2 sin\(\frac {\pi - B}4\) sin\(\frac {\pi - A}4\)
= 1 + 4 sin\(\frac {\pi - A}4\) sin\(\frac {\pi - B}4\) sin\(\frac {\pi - C}4\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = 180°, prove that:
sin2A - sin2B + sin2C = 2 sinA . cosB . sinC
Here,
A + B + C = 180°
or, A + B = 180° - C
sin (A + B) = sin (180° - C) = sinC
cos (A + B) = cos (180° - C) = - cosC
L.H.S.
= sin2A - sin2B + sin2C
= \(\frac {1 - cos 2A}2\) - \(\frac {1 - cos 2B}2\) + sin2C [\(\because\) sin2A = \(\frac {1 - cos 2A}2\)]
= \(\frac 12\) - \(\frac 12\) cos 2A - \(\frac 12\) + \(\frac 12\) cos 2B + sin2C
= \(\frac 12\) cos 2B - \(\frac 12\) cos 2A + sin2C
= \(\frac 12\) (cos 2B + cos 2A) + sin2C
= \(\frac 12\)× 2 sin\(\frac {2B + 2A}2\) sin\(\frac {2A - 2B}2\) + sin2C
= sin (A + B) sin (A - B) + sin2C
= sinC sin (A - B) + sin2C
= sinC [sin (A - B) + sinC]
= sinC [sin (A - B) + sin (A + B)]
= sinC [sinA cosB - cosA sinB + sinA cosB + cosA sinB]
= sinC× 2 sinA cosB
= 2 sinA cosB sinC
Hence, L.H.S. = R.H.S. Proved
If A, B, C are the angles of a triangle, prove that:
sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sinA sinB sinC
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos (\(\pi\) - C) = - cosC
L.H.S.
=sin (B + C - A) + sin (C + A - B) + sin (A + B - C)
= sin (\(\pi\) - A - A) + sin (\(\pi\) - B - B) + sin (\(\pi\) - C - C)
= sin (\(\pi\) - 2A) + sin (\(\pi\) - 2B) + sin (\(\pi\) - 2C)
= 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) + 2 sinC cosC
= 2 sinC . cos (A - B) + 2 sinC cosC
= 2 sinC [cos (A - B) + cosC]
= 2 sinC [cos (A - B) - cos (A + B)]
= 2 sinC [2 sin (\(\frac {A - B + A + B}2\)) . sin (\(\frac {A + B - (A - B)}2\))]
= 2 sinC [2 sin\(\frac {2A}2\) . sin\(\frac {2B}2\)]
= 2 sinC [2 sinA sinB]
= 4 sinA sinB sinC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
\(\frac {sin 2A + sin 2B + sin 2C}{sinA + sinB + sinC}\) = 8 sin\(\frac A2\) . sin\(\frac B2\) . sin\(\frac C2\)
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos (\(\pi\) - C) = - cosC
and
A + B + C = \(\pi\)
or, \(\frac A2\) + \(\frac B2\) + \(\frac C2\) = \(\frac \pi2\)
or, \(\frac A2\) + \(\frac B2\) = \(\frac \pi2\) - \(\frac C2\)
L.H.S.
=\(\frac {sin 2A + sin 2B + sin 2C}{sinA + sinB + sinC}\)
Taking numerator:
sin 2A + sin 2B + sin 2C
= 2 sin (\(\frac {2A + 2B}2\)) . cos (\(\frac {2A - 2B}2\)) + 2 sinC cosC
= 2 sin (A + B) . cos (A - B) + 2 sinC cosC
= 2 sinC . cos (A - B) + 2 sinC cosC
= 2 sinC [cos (A - B) + cosC]
= 2 sinC [cos (A - B) - cos (A + B)]
= 2 sinC [2 sin\(\frac {A + B + A - B}2\) . sin\(\frac {A + B - A + B}2\)]
= 2 sinC [2 sinA . sinB]
= 4 sinA sinB sinC
Taking denominator:
sinA + sinB + sinC
= 2 sin(\(\frac {A + B}2\)) . cos(\(\frac {A - B}2\)) + sinC
= 2 cos\(\frac C2\) .cos(\(\frac {A - B}2\)) + 2 sin\(\frac C2\) cos\(\frac C2\)
=2 cos\(\frac C2\) [cos(\(\frac {A - B}2\)) + sin\(\frac C2\)]
= 2 cos\(\frac C2\) [cos(\(\frac {A - B}2\)) + cos(\(\frac {A + B}2\))]
= 2 cos\(\frac C2\) [2 cos (\(\frac {\frac {A - B}2 + \frac {A + B}2}2\)) - cos (\(\frac {\frac {A - B}2 + \frac {A + B}2}2\))]
= 2 cos\(\frac C2\) [2 cos\(\frac A2\) . cos\(\frac {-B}2\)]
= 2 cos\(\frac C2\) [2 cos\(\frac A2\) . cos\(\frac B2\)]
= 4 cos\(\frac A2\) cos\(\frac B2\) cos\(\frac C2\)
Now,
\(\frac {Numerator}{Denominator}\)
= \(\frac {4 sinA sinB sinC}{4 cos\frac A2 cos\frac B2 cos\frac C2}\)
= \(\frac {2 sin\frac A2 . cos\frac A2 . 2 sin\frac B2 . cos\frac B2 . 2 sin\frac C2 . cos\frac C2}{cos\frac A2 . cos\frac B2 . cos\frac C2}\)
= 8 sin\(\frac A2\) . sin\(\frac B2\) . sin\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), then show that:
cos\(\frac A2\) . cos(\(\frac {B - C}2\)) + cos\(\frac B2\) . cos(\(\frac {C - A}2\)) + cos\(\frac C2\) . cos(\(\frac {A - B}2\)) = sinA + sinB + sinC
Here,
A + B + C = \(\pi\)
or, \(\frac A2\) + \(\frac B2\) + \(\frac C2\) = \(\frac {\pi}2\)
\(\frac A2\) = \(\frac {\pi}2\) - (\(\frac {B + C}2\))
\(\frac B2\) = \(\frac {\pi}2\) - (\(\frac {C + A}2\))
\(\frac C2\) = \(\frac {\pi}2\) - (\(\frac {A + B}2\))
cos\(\frac A2\) = cos (\(\frac {\pi}2\) - (\(\frac {B + C}2\)))
cos\(\frac B2\) = cos (\(\frac {\pi}2\) - (\(\frac {C + A}2\)))
cos\(\frac C2\) = cos (\(\frac {\pi}2\) - (\(\frac {A + B}2\)))
L.H.S.
=cos\(\frac A2\) . cos(\(\frac {B - C}2\)) + cos\(\frac B2\) . cos(\(\frac {C - A}2\)) + cos\(\frac C2\) . cos(\(\frac {A - B}2\))
= cos (\(\frac {\pi}2\) - (\(\frac {B + C}2\))) .cos(\(\frac {B - C}2\)) +cos (\(\frac {\pi}2\) - (\(\frac {C + A}2\))) .cos(\(\frac {C - A}2\)) + cos (\(\frac {\pi}2\) - (\(\frac {A + B}2\))) . cos(\(\frac {A - B}2\))
= \(\frac 12\) [2 sin\(\frac {(B + C)}2\) . cos\(\frac {(B - C)}2\) + 2 sin\(\frac {(C + A)}2\) . cos\(\frac {(C - A)}2\) + 2 sin\(\frac {(A + B)}2\) . cos\(\frac {(A - B)}2\)]
= \(\frac 12\) [sinB + sinC + sinC + sinA + sinA + sinB]
= \(\frac 12\) [2 sinA + 2 sinB + 2 sinC]
= \(\frac 12\)× 2 [sinA + sinB + sinC]
= sinA + sinB + sinC
Hence, L.H.S. = R.H.S Proved
If A + B + C = \(\pi\), prove that:
sin2\(\frac A2\) + sin2\(\frac B2\) + sin2\(\frac C2\) = 1 - 2 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Here,
A + B + C = \(\pi\)
or, A + B = \(\pi\) - C
or, \(\frac A2\) + \(\frac B2\) = \(\frac \pi2\) - \(\frac C2\)
sin (\(\frac A2\) + \(\frac B2\)) = sin (\(\frac \pi2\) - \(\frac C2\)) = cos\(\frac C2\)
cos (\(\frac A2\) + \(\frac B2\)) = cos (\(\frac \pi2\) - \(\frac C2\)) = sin \(\frac C2\)
L.H.S.
=sin2\(\frac A2\) + sin2\(\frac B2\) + sin2\(\frac C2\)
= \(\frac {1 - cosA}2\) + \(\frac {1 + cosB}2\) + sin2\(\frac C2\)
= \(\frac 12\) + \(\frac 12\) - \(\frac 12\) (cosA + cosB) + sin2\(\frac C2\)
= 1 - \(\frac 12\)× 2 cos\(\frac {A + B}2\) cos\(\frac {A - B}2\) + sin2\(\frac C2\)
= 1 - sin\(\frac C2\) cos\(\frac {(A - B)}2\) + sin2\(\frac C2\)
=1 - sin\(\frac C2\) [cos\(\frac {(A - B)}2\) - sin\(\frac C2\)]
=1 - sin\(\frac C2\) [cos\(\frac {(A - B)}2\) - cos\(\frac {(A + B)}2\)]
=1 - sin\(\frac C2\) [2 sin\(\frac {(\frac {A - B}2 + \frac {A + B}2)}2\) sin\(\frac {(\frac {A + B}2 -\frac {A - B}2)}2\)]
=1 - sin\(\frac C2\) [2 sin\(\frac {2A}2\) × \(\frac 12\) sin\(\frac {2B}2\) × \(\frac 12\)]
= 1 - 2 sin\(\frac A2\) sin\(\frac B2\) sin\(\frac C2\)
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\), prove that:
sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin\(\frac {B - C}2\) . sin\(\frac {C - A}2\) . sin\(\frac {A - B}2\)
Here,
A + B + C = \(\pi\)
or, A + B = \(pi\) - C
or, B + C = \(\pi\) - A
or, C + A = \(\pi\) - B
L.H.S.
=sin (B + 2C) + sin (C + 2A) + sin (A + 2B)
= sin (B + C + C) + sin (C + A + A) + sin (A + B + B)
= sin (\(\pi\) - A + C) + sin (\(\pi\) - B + A) + sin (\(\pi\) - C + B)
= sin [\(\pi\) + (C - A)] + sin [\(\pi\) + (A - B)] + sin [\(\pi\) + (B - C)]
= - sin (C - A) - sin (A - B) - sin (B - C)
= - [sin (C - A) + sin (A - B) + sin (B - C)]
= - [2 sin(\(\frac {C - A + A - B}2\)) . cos(\(\frac {C - A - A + B}2\)) + 2 sin(\(\frac {B - C}2\)) . cos(\(\frac {B - C}2\))]
= - [2 sin(\(\frac {C - B}2\)) . cos(\(\frac {B + C - 2A}2\)) + 2 sin(\(\frac {B - C}2\)) . cos(\(\frac {B - C}2\))]
= - [2 sin {-\(\frac {(B - C)}2\)} . cos(\(\frac {B + C - 2A}2\)) + 2 sin(\(\frac {B - C}2\)) . cos(\(\frac {B - C}2\))]
= - [- 2 sin\(\frac {(B - C)}2\) . cos(\(\frac {B + C - 2A}2\)) + 2 sin(\(\frac {B - C}2\)) . cos(\(\frac {B - C}2\))]
= [2 sin\(\frac {(B - C)}2\) . cos(\(\frac {B + C - 2A}2\)) - 2 sin(\(\frac {B - C}2\)) . cos(\(\frac {B - C}2\))]
= 2 sin\(\frac {(B - C)}2\) [cos(\(\frac {B + C - 2A}2\)) - cos(\(\frac {B - C}2\))]
=2 sin\(\frac {(B - C)}2\) [2 sin(\(\frac {\frac {B - C}2 + \frac {B + C - 2A}2}2\)) . sin(\(\frac {\frac {B - C}2 - \frac {B + C - 2A}2}2\))]
=2 sin\(\frac {(B - C)}2\) [2 sin\(\frac {(B - C + B + C - 2A)}4\) . sin\(\frac {(B - C - B - C + 2A)}4\)]
=2 sin\(\frac {(B - C)}2\) [2 sin\(\frac {2(B - A)}4\) . sin\(\frac {2(A - C)}4\)]
=2 sin\(\frac {(B - C)}2\) [2 sin {-\(\frac {(A - B)}2\)} . sin{-\(\frac {(C - A)}2\)}]
=2 sin\(\frac {(B - C)}2\) [2 sin(\(\frac {A - B}2\)) . sin(\(\frac {C - A}2\))]
= 4sin\(\frac {(B - C)}2\) .sin(\(\frac {C - A}2\)) .sin(\(\frac {A - B}2\))
Hence, L.H.S. = R.H.S. Proved
If A + B + C = \(\pi\)c, prove that:
cosC . sinA . cosB + sinB . cosC . cosA + sinC . cosA . cosB = sinA . sinB . sinC
Here,
A + B + C = \(\pi\)c
or, A + B = \(\pi\) - C
sin (A + B) = sin (\(\pi\) - C) = sinC
cos (A + B) = cos (\(\pi\) - C) = -cosC
L.H.S.
=cosC . sinA . cosB + sinB . cosC . cosA + sinC . cosA . cosB
= cosC (sinA cosB + cosA sinB) + sinC . cosA . cosB
= cosC . sin (A + B) + sinC . cosA . cosB
= cosC . sinC + sinC . cosA . cosB
= sinC [cosC + cosA . cosB]
= sinC [- cos (A + B) + cosA . cosB]
= sinC [- cosA . cosB + sinA . sinB + cosA . cosB]
= sinA . sinB . sinC
Hence, L.H.S. = R.H.S. Proved
If A + B + C = 180°, prove that:
\(\frac {sinA}{cosB cosC}\) + \(\frac {sinB}{cosC cosA}\) + \(\frac {sinC}{cosA cosB}\) = 2 tanA tanB tanC
L.H.S.
=\(\frac {sinA}{cosB cosC}\) + \(\frac {sinB}{cosC cosA}\) + \(\frac {sinC}{cosA cosB}\)
= \(\frac {sinA.cosA + sinB.cosB + sinC.cosC}{cosA cosB cosC}\)
= \(\frac {2sinA.cosA + 2sinB.cosB + 2sinC.cosC}{2cosA cosB cosC}\)
= \(\frac {sin 2A + sin 2B + sin2C}{2 cosA cosB cosC}\)
= \(\frac {4 sinA sinB sinC}{2 cosa cosB cosC}\)
= 2 tanA tanB tanC
Hence, L.H.S. = R.H.S. Proved
© 2019-20 Kullabs. All Rights Reserved.