# Conditional Identities

If P, Q, and Q are the angles of a △PQR, prove that:
2$(sin\frac{Q}{2} + sin\frac{R}{2})(sin\frac{Q}{2} - sin\frac{R}{2})$ - cosP = 1 - 4cos$\frac{P}{2}$.cos$\frac{Q}{2}$.sin$\frac{R}{2}$

Given, $\frac P2$ + $\frac Q2$ + $\frac R2$ = $\frac{180°}{2}$ or, $\frac Q2$ + $\frac R2$ = $\frac{180°}{2}$ - $\frac P2$
For cos, cos($\frac Q2$ + $\frac R2$) = cos($\frac{180°}{2}$ - $\frac P2$) = sin$\frac P2$ ---- (i)
For sin, sin($\frac Q2$ + $\frac R2$) = sin($\frac{180°}{2}$ - $\frac P2$) = cos$\frac P2$ ---- (ii)

To prove: 2$(sin\frac{Q}{2} + sin\frac{R}{2})(sin\frac{Q}{2} - sin\frac{R}{2})$ - cosP = 1 - 4cos$\frac{P}{2}$.cos$\frac{Q}{2}$.sin$\frac{R}{2}$

Taking L.H.S,
= 2$(sin\frac{Q}{2} + sin\frac{R}{2})(sin\frac{Q}{2} - sin\frac{R}{2})$ - cosP
= 2$(sin^{2}\frac{Q}{2} - sin^{2}\frac{R}{2})$ - cosP
= 2$(\frac{\text{1 - cosQ}}{2} - \frac{\text{1 - cosR}}{2})$ - cosP
= cosR - cosQ - cosP
= 2$sin\frac{\text{Q + R}}{2}.sin\frac{\text{Q - R}}{2}$ - cos2.$\frac{P}{2}$
= 2cos$\frac{P}{2}.sin\frac{\text{Q - R}}{2}$ - (2cos2$\frac{P}{2}$ - 1) [From 'ii']
= 1 - 2$cos\frac{P}{2}(cos\frac{P}{2} - sin\frac{\text{Q - R}}{2})$
= 1 - 2$cos\frac{P}{2}[sin(\frac Q2 + \frac R2) - sin(\frac{\text{Q - R}}{2})]$ [From 'ii']
= 1 - 2$cos\frac{P}{2}(2cos\frac{Q}{2}.sin\frac R2)$
∴ 1 - 4$cos\frac{P}{2}.cos\frac{Q}{2}.sin\frac R2$ = R.H.S

If A + B + C = 180° or πc then prove that: cos(B + C - A) + cos(C + A - B) + cos(A + B - C) = 1 + cosA.cosB.cosC

Given, A + B + C = 180° ---- (i) or, A + B = 180° - C
For cos, cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: cos(B + C - A) + cos(C + A - B) + cos(A + B - C) = 1 + cosA.cosB.cosC

Taking L.H.S,
= cos(B + C - A) + cos(C + A - B) + cos(A + B - C)
= cos(180° - A - A) + cos(180° - B - B) + cos(180° - C - C) [From (i)]
= cos(180° - 2A) + cos(180° - 2B) + cos(180° - 2C)
= -cos2A - cos2B - cos2C
= -(cos2A + cos2B + cos2C) = -$\left(2cos\frac{\text{2A + 2B}}{2}.cos\frac{\text{2A - 2B}}{2} \text{ + cos2C}\right)$ = -{2cos(A + B)cos(A - B) + cos2C}
= -{-2cosC.cos(A - B) + 2cos2C - 1} [From (ii)]
= 2cosC.cos(A - B) - 2cos2C + 1
= 2cosC {cos(A - B) - cosC} + 1
= 2cosC {cos(A - B) + cos(A + B)} + 1 [Fron (ii)]
= 2cosC.2cosAcosB + 1
∴ 1 + 4 cosA.cosB.cosC = R.H.S

If P + Q + R = 180° then prove that: sin2P + sin2Q + sin2R = 4sinP.sinQ.sinR

Given, P + Q + R = 180° or, P + Q = 180° - R
For sin, sin(P + Q) = sin(180° - R) = sinR ---- (i)
For cos, cos(P + Q) = cos(180° - R) = -cosR ---- (ii)

To prove: sin2P + sin2Q + sin2R = 4sinP.sinQ.sinR

Taking L.H.S,
= sin2P + sin2Q + sin2R
= 2sin$\left(\frac{\text{2P + 2Q}}{2}\right)$cos$\left(\frac{\text{2P - 2Q}}{2}\right)$ + sin2R
= 2sin(P + Q).cos(P - Q) + sin2R
= 2sinR.cos(P- Q) + 2sinRcosR [From (i)]
= 2sinR{cos(P - Q) + cosR}
= 2sinR{cos(P - Q) - cos(P + Q)} [From (ii)]
= 2sinR.2sinPsinQ
∴ 4sinP.sinQ.sinR = R.H.S

If A + B + C = 180° or πc then prove that: sinA - sinB + sinC = 4sin$\frac A2$cos$\frac B2$sin$\frac C2$

Here, $\frac A2$ + $\frac B2$ + $\frac C2$ = $\frac{180°}{2}$ or, $\frac A2$ + $\frac B2$ = $\frac{180°}{2}$ - $\frac C2$
For cos, cos($\frac A2$ + $\frac B2$) = cos($\frac{180°}{2}$ - $\frac C2$) = sin$\frac C2$ ---- (i)
For sin, sin($\frac A2$ + $\frac B2$) = sin($\frac{180°}{2}$ - $\frac C2$) = cos$\frac C2$ ---- (ii)

To prove: sinA - sinB + sinC = 4sin$\frac A2$cos$\frac B2$sin$\frac C2$

Taking L.H.S,
= sinA - sinB + sinC
= 2cos$\frac{\text{A + B}}{2}$.sin$\frac{\text{A - B}}{2}$ + sin2.$\frac C2$

= 2$sin\frac C2$.sin$\frac{\text{A - B}}{2}$ + 2sin$\frac C2$.cos$\frac C2$ [From (i)]
= 2sin$\frac C2 \left(sin\frac{\text{A - B}}{2} \text{ + }cos\frac C2\right)$ = 2sin$\frac C2 \{sin\left(\frac A2 \text{ - }\frac B2\right) \text{ + }sin\left(\frac A2 \text{ + }\frac B2\right)\}$ [From (ii)] = 2sin$\frac C2$.2sin$\frac A2$cos$\frac B2$
∴ 4sin$\frac A2$cos$\frac B2$sin$\frac C2$ = R.H.S

If A + B + C = 180° or πc then prove that: cos2A + cos2B + 2cosAcosBcosC = sin2C.

Here, A + B + C = 180° or, A + B = 180° - C
For cos, cos(A + B) = cos(180° - C) = -cosC ---- (i)

To prove: cos2A + cos2B + 2cosAcosBcosC = sin2C

Taking L.H.S,
= cos2A + cos2B + 2cosAcosBcosC
= $\frac 12$(2cos2A + 2cos2B + 4cosA.cosB.cosC)
= $\frac 12$(1 + cos2A + 1 + cos2B + 4cosA.cosB.cosC)
= $\frac 12$ $\{$2cos$\frac{\text{2A + 2B}}{2}$.cos$\frac{\text{2A - 2B}}{2}$ + 2 + 4cosA.cosB.cosC$\}$

= $\frac 12${2cos(A + B).cos(A - B) + 2 + 4cosA.cosB.cosC}
= cos(A + B).cos(A - B) + 1 + 2cosA.cosB.cosC
= -cosC.cos(A - B) + 1 + 2cosA.cosB.cosC [From (i)]
= cosC {2cosA.cosB - cos(A - B)} + 1
= cosC {cos(A + B) + cos(A - B) - cos(A - B)} + 1
= cosC {cos(A + B)} + 1
= cosC(-cosC) + 1 [From (i)]
= -cos2 + 1
∴ sin2C = R.H.S

If P + Q + R = 180° then prove that: sinP + sinQ + sinR = 4cos$\frac P2$.cos$\frac Q2$.cos$\frac R2$

Given, $\frac P2$ + $\frac Q2$ + $\frac R2$ = 180° or, $\frac P2$ + $\frac Q2$ = $\frac{180°}{2}$ - $\frac R2$
For sin, sin($\frac P2$ + $\frac Q2$) = sin($\frac{180°}{2}$ - $\frac R2$) = cos$\frac R2$ ---- (i)
For cos, cos($\frac P2$ + $\frac Q2$) = cos($\frac{180°}{2}$ - $\frac R2$) = sin$\frac R2$ ---- (ii)

To prove: sinP + sinQ + sinR = 4cos$\frac P2$.cos$\frac Q2$.cos$\frac R2$

Taking L.H.S,
= sinP + sinQ + sinR
= 2sin$\left(\frac{\text{P + Q}}{2}\right)$cos$\left(\frac{\text{P - Q}}{2}\right)$ + sin2.$\frac R2$
= 2cos$\frac R2$.cos$\left(\frac{\text{P - Q}}{2}\right)$ + 2sin$\frac R2$cos$\frac R2$ [From (i)]
= 2cos$\frac R2$ $\{cos\left(\frac{\text{P - Q}}{2}\right) \text{ + }cos\left(\frac{\text{P + Q}}{2}\right)\}$ [From (ii)]
= 2cos$\frac R2$.2cos$\frac P2$cos$\frac Q2$
∴ 4cos$\frac P2$.cos$\frac Q2$.cos$\frac R2$ = R.H.S

If A + B + C = 180° or πc then prove that: $\frac{cosA}{sinB.sinC}$ + $\frac{cosB}{sinC.sinA}$ + $\frac{cosC}{sinA.sinB}$ = 2

Here, A + B + C = 180° or, A + B = 180° - C
For sin, sin(A + B) = sin(180° - C) = sinC ---- (i)
For cos, cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: $\frac{cosA}{sinB.sinC}$ + $\frac{cosB}{sinC.sinA}$ + $\frac{cosC}{sinA.sinB}$ = 2

Taking L.H.S,
= $\frac{cosA}{sinB.sinC}$ + $\frac{cosB}{sinC.sinA}$ + $\frac{cosC}{sinA.sinB}$
= $\frac{\text{sinA.cosA + sinB.cosB + sinC.cosC}}{sinA.sinB.sinC}$
= $\frac{\text{sin2A + sin2B + sin2C}}{2sinA.sinB.sinC}$
= $\frac{2sin\left(\frac{\text{2A + 2B}}{2}\right).cos\left(\frac{\text{2A - 2B}}{2}\right) \text{ + sin2C}}{2sinA.sinB.sinC}$
= $\frac{\text{2sin(A + B).cos(A - B) + sin2C}}{2sinA.sinB.sinC}$
= $\frac{\text{2sinC.cos(A - B) + 2sinC.cosC}}{2sinA.sinB.sinC}$ [From (i)]
= $\frac{2sinC\{\text{cos(A - B) - cos(A + B)}\}}{2sinA.sinB.sinC}$ [From (ii)]

= $\frac{2sinC.2sinAsinB}{2sinA.sinB.sinC}$
= 2

If A + B + C = 180° or πc then prove that: sin2A + sin2B + sin2C = 2 + 2cosA.cosB.cosC

Here, A + B + C = 180° or, A + B = 180° - C
For sin, sin(A + B) = sin(180° - C) = sinC ---- (i)
For cos, cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: sin2A + sin2B + sin2C = 2 + 2cosA.cosB.cosC

Taking L.H.S,
= sin2A + sin2B + sin2C
= $\frac 12$(2sin2A + 2sin2B + 2sin2C)
= $\frac 12$(1 - cos2A + 1 - cos2B + 1 - cos2C)
= $\frac 12${3 - (cos2A + cos2B) - cos2C}
= $\frac 12$ $\{$ 3 - 2cos$\left(\frac{\text{2A + 2B}}{2}\right)$.cos$\left(\frac{\text{2A - 2B}}{2}\right)$ - cos2C$\}$
= $\frac 12${3 - 2cos(A + B).cos(A - B) - cos2C}
= $\frac 12${3 - 2(-cosC)cos(A - B) - cos2C} [From (ii)]
= $\frac 12${3 + 2cosC.cos(A - B) - 2(cos2C - 1)}
= $\frac 12${3 + 2cosC.cos(A - B) - 2cos2C + 1}
= $\frac 12$[4 + 2cosC{cos(A - B) - cosC}]

= $\frac 12$ [4 + 2cosC{cos(A - B) + cos(A + B)}] [From (ii)]
= 2 + cosC.2cosA.cosB
∴ 2 + 2cosA.cosB.cosC = R.H.S

If α + β + θ = π, prove that: sin(α + β - θ) + sin(β + θ - α) + sin(θ + α - β) = 4sinα.sinβ.sinθ

Here, α + β + θ = π ---- (i) or, α + β = π - θ
For sin, sin(α + β) = sin(π - θ) = sinθ ---- (ii)
For cos, cos(α + β) = cos(π - θ) = -cosθ ---- (iii)

To prove: sin(α + β - θ) + sin(β + θ - α) + sin(θ + α - β) = 4sinα.sinβ.sinθ

Taking L.H.S,
= sin(α + β - θ) + sin(β + θ - α) + sin(θ + α - β)
= sin(π - θ - θ) + sin(π - α - α) + sin(π - β - β) [From (i)]
= sin(π - 2θ) + sin(π - 2α) + sin(π - 2β)
= sin2θ + sin2α + sin2β
= 2sin$\left(\frac{\text{2α + 2β}}{2}\right)$.cos$\left(\frac{\text{2α - 2β}}{2}\right)$ + sin2θ
= 2sin(α + β).cos(α - β) + sin2θ
= 2sinθ.cos(α - β) + 2sinθ.cosθ [From (ii)]
= 2sinθ{cos(α - β) + cosθ}
= 2sinθ{cos(α - β) - cos(α + β)} [From (iii)]
= 2sinθ.2sinα.sinβ
∴ 4sinθ.sinα.sinβ = R.H.S

If A + B + C = 180° or πc then prove that: $\frac{\text{sin2A + sin2B + sin2C}}{4cos\frac A2.cos\frac B2.cos\frac C2}$ = 8sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Here, A + B + C = 180° or, A + B = 180° - C
For sin, sin(A + B) = sin(180° - C) = sinC ---- (i)
For cos, cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: $\frac{\text{sin2A + sin2B + sin2C}}{4cos\frac A2.cos\frac B2.cos\frac C2}$ = 8sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Taking L.H.S,
= $\frac{\text{sin2A + sin2B + sin2C}}{4cos\frac A2.cos\frac B2.cos\frac C2}$ ---- (A)

Firstly,
= sin2A + sin2B + sin2C
= 2sin$\left(\frac{\text{2A + 2B}}{2}\right)$.cos$\left(\frac{\text{2A - 2B}}{2}\right)$ + sin2C
= 2sin(A + B).cos(A - B) + sin2C
= 2sinC.cos(A - B) + 2sinC.cosC [From (i)]
= 2sinC{cos(A - B) + cosC}
= 2sinC{cos(A - B) + cos(A + B)} [From (ii)]
= 2sinC.2sinA.sinB ---- (B)

Finally, putting the value of (B) in (A), we get,
= $\frac{4sinC.sinA.sinB}{4cos\frac A2.cos\frac B2.cos\frac C2}$
= $\frac{sin2.\frac C2sin2.\frac A2sin2.\frac B2}{cos\frac A2.cos\frac B2.cos\frac C2}$
= $\frac{2sin\frac C2 cos\frac C2 \times 2sin\frac A2 cos\frac A2 \times 2sin\frac B2 cos\frac B2}{cos\frac A2.cos\frac B2.cos\frac C2}$
∴ 8sin$\frac A2$.sin$\frac B2$.sin$\frac C2$ = R.H.S

If A + B + C = 180° or πc then prove that: cosA + cosB + cosC = 1 + 4sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Given, $\frac A2$ + $\frac B2$ + $\frac C2$ = 180° $\frac A2$ + $\frac B2$ = $\frac{180°}{2}$ - $\frac C2$
cos($\frac A2$ + $\frac B2$) = cos($\frac{180°}{2}$ - $\frac C2$) = sin$\frac C2$ ---- (i)
sin($\frac A2$ + $\frac B2$) = sin($\frac{180°}{2}$ - $\frac C2$) = cos$\frac C2$ ---- (ii)

To prove: cosA + cosB + cosC = 1 + 4sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Taking L.H.S, = cosA + cosB + cosC
= 2cos$\frac{\text{A + B}}{2}$cos$\frac{\text{A - B}}{2}$ + cos2.$\frac C2$
= 2sin$\frac C2$cos$\frac{\text{A - B}}{2}$ + 1 - 2sin2$\frac C2$ [From (i)]
= 2sin$\frac C2$ $\left(cos\frac{\text{A - B}}{2}\text{ - }sin\frac C2\right)$ + 1
= 2sin$\frac C2$ $\left(cos\frac{\text{A - B}}{2}\text{ - }cos\frac{\text{A + B}}{2}\right)$ + 1 [From (i)]
= 1 + 2sin$\frac C2$.2sin$\frac A2$.sin$\frac B2$
∴ 1 + 4sin$\frac A2$.sin$\frac B2$.sin$\frac C2$ = R.H.S

If A + B + C = 180° or πc then prove that: sin2$\frac A2$ + sin2$\frac B2$ + sin2$\frac C2$ = 1 - 2sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Here, A + B + C = 180° $\frac A2$ + $\frac B2$ + $\frac C2$ = $\frac{180°}{2}$
$\frac A2$ + $\frac B2$ = $\frac{180°}{2}$ - $\frac C2$
cos($\frac A2$ + $\frac B2$) = cos($\frac{180°}{2}$ - $\frac C2$) = sin$\frac C2$ ---- (i)
sin($\frac A2$ + $\frac B2$) = sin($\frac{180°}{2}$ - $\frac C2$) = cos$\frac C2$ ---- (ii)

To prove: sin2$\frac A2$ + sin2$\frac B2$ + sin2$\frac C2$ = 1 - 2sin$\frac A2$.sin$\frac B2$.sin$\frac C2$

Taking L.H.S,
= sin2$\frac A2$ + sin2$\frac B2$ + sin2$\frac C2$
= $\frac 12 \left(2sin^{2}\frac A2 \text{ + }2sin^{2}\frac B2\text{ + }2sin^{2}\frac C2\right)$
= $\frac 12 \left(\text{1 - }cos2.\frac A2 \text{ + 1 - }cos2.\frac B2 \text{ + 1 - }cos2.\frac C2\right)$
= $\frac 12 \{\text{3 - (cosA + cosB) - cosC}\}$
= $\frac 12 \{\text{3 - }2cos\left(\frac{\text{A + B}}{2}\right)cos\left(\frac{\text{A - B}}{2}\right) \text{ - }cos2.\frac C2\}$
= $\frac 12 \{\text{3 - }2sin\frac C2 cos\left(\frac{\text{A - B}}{2}\right)\text{ - (1 - }2sin^{2}\frac C2)\}$
= $\frac 12 \{\text{3 - }2sin\frac C2 cos\left(\frac{\text{A - B}}{2}\right)\text{ - 1 + }2sin^{2}\frac C2\}$
= $\frac 12 \{\text{2 + }2sin^{2}\frac C2 \text{ - }2sin\frac C2 cos\left(\frac{\text{A - B}}{2}\right)\}$
= $\frac 22 [\text{1 + }sin\frac C2\{sin\frac C2 \text{ - }cos\left(\frac{\text{A - B}}{2}\right)\}]$
= $\text{1 + }sin\frac C2\{cos\left(\frac{\text{A + B}}{2}\right) \text{ - }cos\left(\frac{\text{A - B}}{2}\right)\}$ [From (i)]
= $\text{1 - }sin\frac C2\{cos\left(\frac{\text{A - B}}{2}\right) \text{ - }cos\left(\frac{\text{A + B}}{2}\right)\}$
= 1 - sin$\frac C2$.2sin$\frac A2$.sin$\frac B2$
∴ 1 - 2sin$\frac A2$.sin$\frac B2$.sin$\frac C2$ = R.H.S

If X + Y + Z = 180° then prove that: cos2X + cos2Y - cos2Z = 1 - 4sinX.sinY.cosZ

Here, X + Y + Z = 180° X + Y = 180° - Z
sin(X + Y) = sin(180° - Z) = sinZ ---- (i)
cos(X + Y) = cos(180° - Z) = -cosZ ---- (ii)

To prove: cos2X + cos2Y - cos2Z = 1 - 4sinX.sinY.cosZ

= cos2X + cos2Y - cos2Z
= 2cos$\left(\frac{\text{2X + 2Y}}{2}\right).cos\left(\frac{\text{2X - 2Y}}{2}\right)$ - cos2Z
= 2cos(X + Y).cos(X - Y) - cos2Z
= 2(-cosZ).cos(X -Y) - (2cos2Z - 1) [From (ii)]
= -2cosZ.cos(X - Y) - 2cos2Z + 1
= -2cosZ{cos(X- Y) + cosZ} + 1
= -2cosZ{cos(X - Y) - cos(X + Y)} + 1 [From (ii)]
= -2cosZ.2sinX.sinY + 1
∴ 1 - 4sinXsinYcosZ = R.H.S

If A + B + C = 180° or πc then prove that: sin2$\frac A2$ + sin2$\frac B2$ - sin2$\frac C2$ = 1 - 2cos$\frac A2$.cos$\frac B2$.sin$\frac C2$

Here, A + B + C = 180° $\frac A2$ + $\frac B2$ + $\frac C2$ = $\frac{180°}{2}$
$\frac A2$ + $\frac B2$ = $\frac{180°}{2}$ - $\frac C2$
sin($\frac A2$ + $\frac B2$) = sin($\frac{180°}{2}$ - $\frac C2$) = cos$\frac C2$ ---- (i)
cos($\frac A2$ + $\frac B2$) = cos($\frac{180°}{2}$ - $\frac C2$) = sin$\frac C2$ ---- (ii)

To prove: sin2$\frac A2$ + sin2$\frac B2$ - sin2$\frac C2$ = 1 - 2cos$\frac A2$.cos$\frac B2$.sin$\frac C2$

Taking L.H.S,
= sin2$\frac A2$ + sin2$\frac B2$ - sin2$\frac C2$
= $\frac 12 \left(2sin^{2}\frac A2 \text{ + }2sin^{2}\frac B2 \text{ - }2sin^{2}\frac C2\right)$
= $\frac 12 \{\text{1 - }cos2.\frac A2 \text{ + 1 -}cos2.\frac B2 \text{ -(1 - }cos2.\frac C2)\}$
= $\frac 12 \left(\text{1 - cosA + 1 - cosB - 1 + cosC}\right)$
= $\frac 12 \{\text{1 - (cosA + cosB) + cosC}\}$
= $\frac 12 \{\text{1 - }2cos\left(\frac{\text{A + B}}{2}\right).cos\left(\frac{\text{A - B}}{2}\right)\text{ + }cos2.\frac C2\}$
= $\frac 12 \{\text{1 - }2sin\frac C2.cos\left(\frac{\text{A - B}}{2}\right)\text{ + 1 - }2sin^{2}\frac C2\}$ [From (ii)]
= $\frac 12 [\text{2 - }2sin\frac C2 \{cos\left(\frac{\text{A - B}}{2}\right) \text{ + }sin\frac C2\}]$
= $\frac 22 [\text{1 - }sin\frac C2 \{cos\left(\frac{\text{A - B}}{2}\right) \text{ + }cos\left(\frac{\text{A + B}}{2}\right)\}]$ [From (ii)]
= 1 - sin$\frac C2$.2cos$\frac A2$.cos$\frac B2$
∴ 1 - 2cos$\frac A2$.cos$\frac B2$.sin$\frac C2$ = R.H.S

If A + B + C = 180° or πc then prove that: sin(B + C - A) + sin(C + A - B) + sin(A + B - C) = 4sinA.sinB.sinC

Here, A + B + C = 180° ---- (i) A + B = 180° - C
sin(A + B) = sin(180° - C) = sinC ---- (ii)
cos(A + B) = cos(180° - C) = -cosC ---- (iii)

To prove: sin(B + C - A) + sin(C + A - B) + sin(A + B - C) = 4sinA.sinB.sinC

Taking L.H.S,
= sin(B + C - A) + sin(C + A - B) + sin(A + B - C)
= sin(180° - A - A) + sin(180° - B - B) + sin(180° - C - C) [From (i)]
= sin(180° - 2A) + sin(180° - 2B) + sin(180° - 2C)
= sin2A + sin2B + sin2C
= 2sin$\left(\frac{\text{2A + 2B}}{2}\right)$.cos$\left(\frac{\text{2A - 2B}}{2}\right)$ + sin2C
= 2sin(A + B).cos(A - B) + sin2C
= 2sinC.cos(A - B) + 2sinC.cosC [From (ii)]
= 2sinC {cos(A - B) + cosC}
= 2sinC {cos(A - B) - cos(A + B)} [From (iii)]
= 2sinC.2sinA.sinB
∴ 4sinA.sinB.sinC = R.H.S

If A + B + C = 180° or πc then prove that: cos2A + cos2B + cos2C = -1 - 4cosA.cosB.cosC

Here, A + B + C = 180° A + B = 180° - C
sin(A + B) = sin(180° - C) = sinC ---- (i)
cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: cos2A + cos2B + cos2C = -1 - 4cosA.cosB.cosC

Taking L.H.S,
= 2cos$\left(\frac{\text{2A + 2B}}{2}\right)$cos$\left(\frac{\text{2A - 2B}}{2}\right)$ + cos2C
= 2cos(A + B).cos(A - B) + cos2C
= -2cosC.cos(A - B) + 2cos2C - 1
= -2cosC{cos(A - B) - cosC} - 1
= -2cosC{cos(A - B) + cos(A + B)} - 1 [From (ii)]
= -2cosC.2cosA.cosB - 1
∴ -4cosA.cosB.cosC - 1 = R.H.S

If A + B + C = 180° or πc then prove that: sin2A - sin2B + sin2C = 2sinA.cosB.sinC

Here, A + B + C = 180° A + B = 180° - C
sin(A + B) = sin(180° - C) = sinC ---- (i)
cos(A + B) = cos(180° - C) = -cosC ---- (ii)

To prove: sin2A - sin2B + sin2C = 2sinA.cosB.sinC

Taking L.H.S,
= sin2A - sin2B + sin2C
= $\frac 12$(2sin2A - 2sin2B + 2sin2C)
= $\frac 12$(1 - cos2A - 1 + cos2B + 1 - cos2C)
= $\frac 12$(1 - cos2A + cos2B - cos2C)
= $\frac 12$(1 + cos2B - cos2A - cos2C) = $\frac 12 \{\text{1 + }2sin\left(\frac{\text{2A + 2B}}{2}\right)sin\left(\frac{\text{2A - 2B}}{2}\right)\text{ - cos2C}\}$ = $\frac 12$ {1 + 2sin(A + B).sin(A - B) - cos2C}
= $\frac 12$ {1 + 2sinC.sin(A - B) - (1 - 2sin2C)} [From (i)]
= $\frac 12$ {1 + 2sinC.sin(A - B) - 1 + 2sin2C}
= $\frac 12$ [2sinC{sin(A - B) + sinC}]
= $\frac 12$ [2sinC{sin(A - B) + sin(A + B)}] [From (i)]
= $\frac 12$(2sinC.2sinA.cosB)
∴ 2sinA.cosB.sinC = R.H.S