Conditional Trigonometric Identities

Conditional Trigonometric Identities.

examples for 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 \(\frac{A}{2}\)

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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 \(\frac{A}{2}\)

[removed][removed]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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