SEE OPT Maths: Conditional Trigonometric Identities

In conditional trigonometric identities we will discuss certain relationship exists among the angles involved. We know some of the trigonometric identities which were true for all values of the angles involved. These identities hold for all values of the angles which satisfy the given conditions among them and hence they are called conditional trigonometric identities.

Such identities involving different trigonometrical ratios of three or more angles can be deduced when these angles are connected by some given relation. Suppose, if the sum of three angles be equal to two right angles then we can establish many important identities involving trigonometrical ratios of those angles. To establish such identities we require to use the properties of supplementary and complementary angles.

If A, B and C denote the angles of a triangle ABC, then the relation A + B + C = π enables us to establish many important identities involving trigonometric ratios of these angles The following results are useful to obtain the said identities.

If A + B + C = π, then the sum of any two angles is supplementary to the third i.e.,

(i) B + C = π - A or, C + A = π - B or A + B = π - C. 

(ii) If A + B + C = π then 

sin (A + B) = sin (π - C) = sin C 

sin (B + C) = sin (π - A) = sin A 

  sin (C + A) = sin (π - B) = sin B

(iii) If A + B + C = π then 

cos (A + B) = cos (π - C) = - cos C  
cos (B + C) = cos (π - A) = - cos A
cos (C + A) = cos (π - B) = - cos B

(iv) If A + B + C = π then 

tan (A + B) = tan (π - C) = - tan C 

tan (B + C) = tan (π - A) = - tan A 

tan (C + A) = tan (π - B) = - tan B

(v) If A + B + C = π  then, 

A2 + B2 + C2 = π2 

Hence, it is evident that the sum of any two of the three angles 

C2, B2, A/2
 is complementary to the third. 

i.e., A+B2 = ππ/2 - CC/2,

B+C2$\frac{ }{}$ = π2$\frac{ }{}$ - A2$\frac{ }{}$

C+A2$\frac{ }{}$ = π2$\frac{ }{}$ - B2$\frac{ }{}$

 

Therefore,

sin (A2$\frac{ }{}$ + B2$\frac{ }{}$) = sin π2$\frac{ }{}$ - C2$\frac{ }{}$ = cos C2$\frac{ }{}$

sin (B2$\frac{ }{}$ + C2$\frac{ }{}$) = sin π2$\frac{ }{}$ - A2$\frac{ }{}$ = cos A2$\frac{ }{}$

sin (C2$\frac{ }{}$ + A2$\frac{ }{}$) = sin π2$\frac{ }{}$ - B2$\frac{ }{}$ = cos B2$\frac{ }{}$

cos (A2$\frac{ }{}$ + B2$\frac{ }{}$) = cos π2$\frac{ }{}$ - C2$\frac{ }{}$ = sin C2$\frac{ }{}$

sin (B2$\frac{ }{}$ + C2$\frac{ }{}$) = cos π2$\frac{ }{}$ - A2$\frac{ }{}$ = sin A2$\frac{ }{}$

sin (C2$\frac{ }{}$ + A2$\frac{ }{}$) = cos π2$\frac{ }{}$ - B2$\frac{ }{}$ = sin B2$\frac{ }{}$

tan (A2$\frac{ }{}$ + B2$\frac{ }{}$) = tan π2$\frac{ }{}$ - C2$\frac{ }{}$ = cot C2$\frac{ }{}$

tan (B2$\frac{ }{}$ + C2$\frac{ }{}$) = tan π2$\frac{ }{}$ - A2$\frac{\mathrm{ }}{}$= cot A2$\frac{ }{}$

tan (C2$\frac{ }{}$ + A2$\frac{ }{}$) = tan π2$\frac{ }{}$ - B2$\frac{ }{}$ = cot B2$\frac{ }{}$

In conditional trigonometric identities we will discuss certain relationship exists among the angles involved. We know some of the trigonometric identities which were true for all values of the angles involved. These identities hold for all values of the angles which satisfy the given conditions among them and hence they are called conditional trigonometric identities.

Proof of some conditional trigonometric identities.

1. If A + B + C = 180° then prove that:

cos(B + C – A) + cos(C + A – B) + cos(A + B + C) = 1 + 4 cosA.cosB.cosC

Proof: Here, 
      A + B + C = πc
i.e. A + B = πc - C 
 ⇒    sin(A + B) = sin(πc) sinC 
 ⇒    cos(A +  B) = cos (πc - C) = -cosC.

LHS = cos(B + C – A) + cos(C + A – B) + cos(A + B - C) 
         = cos(180° - 2A) + cos(180° - 2B) + cos(180° - 2C)
         = - cos2A + cos2B – cos2C
         = - (cos2A + cos2B) – cos2C
         = - 2 cos(A + B) cos(A – B) – cos2C
         = 2 cosC cos(A – B) – 2cos² C + 1
         = 1 + 2 cosC [cos (A – B) – cosC] 
         = 1 + 2cosC [cos(A – B) + cos(A + B)] 
         = 1 + 2cosC (2cosA cosB)
         = 1 + 4cosA cosB cosC
         = RHS.

Proved.

 2. If A + B + C = 180° then prove that:

cos(B + C – A) + cos(C + A – B) + cos(A + B + C) = 1 + 4 cosA.cosB.cosC 

Proof: Here, 
      A + B + C = πc
⇒   A + B = π^c- C
  ⇒     sin(A + B) = sin(π ^c - C) sinC
 ⇒        cos(A + B) = cos (π^c - C) = -cosC.

LHS = cos(B + C – A) + cos(C + A – B) + cos(A + B - C)
         = cos(180° - 2A) + cos(180° - 2B) + cos(180° - 2C)
         = - cos2A + cos2B – cos2C
         = - (cos2A + cos2B) – cos2C
         = - 2 cos(A + B) cos(A – B) – cos2C
         = 2 cosC cos(A – B) – 2cos² C + 1
         = 1 + 2 cosC [cos (A – B) – cosC]
         = 1 + 2cosC [cos(A – B) + cos(A + B)]
         = 1 + 2cosC (2cosA cosB)
         = 1 + 4cosA cosB cosC
         = RHS.

Proved.

3. If P + Q + R = 180°  then prove that:

sin2P + sin 2Q + sin 2R = 4 sin P. sin Q. sin R

Proof: Here,
      P + Q + R = π                     
⇒ P + Q = π - R  
Operating by sin & cos, 
     sin (P + Q) =  sin R            
     cos (P + Q) = - cos R.

LHS = sin 2P + sin2Q + sin 2R    
         = 2 sin 2P+2Q2 cos 2P-2Q2 + sin 2R
         = 2 sin (P + Q) cos (P - Q) + sin 2R
         = 2 sin R cos (P - Q) + 2sin R cos R
         = 2 sin R {cos(P - Q) + cos R}
         = 2 sin R{ cos (P- Q) - cos (P + Q)}
         = 2 sin R. 2 cos P – Q +P + Q2 sin P + Q –P + Q2
         = 4 sin R sin P sin Q
         = 4 sin P sin Q sin R
         = RHS.

Proved.

4. If A + B + C =180° , Prove that:
sin A - sin B + sin C = 4 sin A2 cos B2 sin C2  . 
Proof: Here,
      A + B + C = π
⇒    A2 + B2 +C2 = π2
⇒   A2 + B2 =π2-C2 
∴ sin(A2 + B2) = cos C2 and ⁡cos(A2 + B2) = ⁡sinC2 

LHS = sin A - sin B + sin C
         = 2⁡cosA+B2⁡sin A-B2+sin C
         = 2⁡sinC2⁡sinA-B2+ 2 sin C2 ⁡cosC2
         = 2 sinC2(sinA-B2 +⁡cos C2)
         = 2 sinC2(sinA-B2 + sinA+B2)
     = 2 sin C2×2 sin  A-B2+A+B22  cos A-B2 – A+B22
         = 4 sin C2sinA-B+A+B4 +⁡cos A-B-A-B4
         = 4 sin C2 sin A2 cos B2
         = 4 sin A2 cos B2 sin C2  .
         =  RHS.

Proved.

5. If A + B + C =180° then Prove that: cos² A+ cos² B +2cos A cos B cos C = sin² C. 
Proof: Here, 
     A + B =180°- C 
∴  cos (A + B) =  cos (180°-C) =  - cos c     
& sin (A + B) =  sin (180°-C) =  sin c.
LHS = cos² A+ cos² B +2cos A cos B cos C
         = 1+cos2A2 + 1-cos2B2 + 2cos A cos B cos C
         = 12 + 12cos2A +12 + 12 cos2B + 2cosA cosB cosC
         = 1 + 12 (cos 2A + cos2B ) + 2cosA cosB cosC
         = 1 + 12 . 2cos (2A+2B2)cos(2A-2B2)+ 2cos A cos B cos C
      = 1+ cos (A + B) cos (A - B) + 2cosA cosB cosC
         = 1 - cos C cos (A - B) + 2cos A cos B cos C
         = 1 - cos C [cos (A – B) – 2cosA cosB] 
         = 1 – cosC [cosA cosB + sinA sinB – 2cosA cosB]
         = 1 + cosC [cosA cosB – sinA sinB ] 
         = 1 + cosC cos (A + B) 
         = 1 + cosC (- cos C )
         = 1 - cos² C 
         = sin²C 
         = RHS.

Proved. 

6. If P + Q + R = π^c then prove that: 

sin P + sin Q + sin R = 4 cos   P2.cos Q2.cos R2 

Proof: Here, 
      P + Q + R = 180° 
⇒    P2+ Q2+ R2=180°2
⇒    P2+ Q2=90° - R2
⇒    sin (P2+ Q2)=sin (90° -R2)
∴  sin (P2+ Q2)= cos R2

RHS = sin P + sin Q + sin R
         = 2sin P+Q2.. cosP-Q2 + 2sin R2.cos R2
         = 2cosR2 ( cos P-Q2 + sin R2)
         = 2cosR2 ( cos P-Q2+ cos P + Q2) [∵ sin P+Q2=cos R2]
         = 2 cos R2. 2 cos P-Q2 + P+Q22 . cos P-Q2- P+Q22
         = 4 cos R2 cos P2cos(-Q2)
         = 4 cos P2cos Q2 cos R2
         =RHS.

 Proved.

7. If P + Q + R = 180° then, prove that:

   cosPsin Q.sinR+ cosQsin R.sinP+cosRsin P.sinQ= 2

Proof: Here,
      P + Q + R =180°
⇒    P + Q =180°- R

Taking sin and cos on both the sides then, 
        sin (P + Q) = sin (180°- R) =   sin R
&     cos (P + Q) = cos (180°- R) = - cos R

LHS = cosPsin Q. ⁡sinR+ ⁡cosQsin R.⁡sinP+⁡cosRsin P.⁡sinQ
        = ⁡cosP ⁡sinP + ⁡cosQ sin Q+⁡cosRsin⁡sinRsin P. sinQ. . sinR
        = ⁡2cosP ⁡sinP +⁡2cosQ sin Q+⁡2Rsin⁡sinR2sin P.⁡ sinQ. sinR
        = sin 2P +sin 2Q+ ⁡ ⁡sin2R2sin P⁡sinQ sinR......... (A)

Taking numerator only,
    sin2P + sin 2Q + sin 2R
= 2 sin (2P+2Q2cos2P-2Q2)+ 2 sin R cos R
= 2 sin (P +Q) cos (P-Q) + 2 sin R cos R
= 2 sin R cos (P-Q) + 2 sin R cos R
= 2 sin R [cos (P-Q) +  cos R]
= 2 sin R [cos (P-Q) -  cos (P+Q)]
= 2 sin R. 2 sin P sin Q
= 4 sin R sinQ sin R

Now, relation (A) becomes;
    4⁡sinP sinQ⁡sinR    2⁡sinP sinQ⁡sinR
  = 2.
 = RHS.

Proved.

8. If A + B + C =180° , Prove that:
sin² A+ sin² B + sin ² C = 2 + 2cos A. cos B.cos C
Proof:  Here, 
      A + B + C =180°
⇒  A + B =180°- C .

Taking sin and cos on both the sides
       sin (A + B) =  sin (180°-C) =  sin C     
&   cos (A + B) =  cos (180°-C) =  - cos C

LHS =   sin² A+ sin² B + sin ² C

        = 1- cos2A2 + 1-cos2B2 + sin² C
        = 12 - 12 cos2A+ 12 - 12 cos2B + sin ² C
        = 1 - 12 (cos2A+ cos2B) + sin ² C
        = 1 - 122cos2A+2B2 cos 2A-2B2 + sin ² C
        = 1 - cos (A+B) cos (A-B) + sin ² C
        = 1 + cos C cos (A-B) + 1- cos² C

 (∵cos(A+B)= -cosC)

        = 2 + cos C [cos (A-B) - cos C]
        = 2 + cos C [cos (A-B) + cos (A + B)]
        = 2 + 2cos A cos B cos C
        = RHS.

Proved.

9. If A + B + C = π^c , prove that: 

cosAsin B.sinC+cosBsin C.sinA+cosCsin A.sinB = 2

Proof: do yourself.

10. If A + B + C = π^c , prove that: 
 sin2A+sin2B+sin2C4⁡cos A2.⁡cos B2.cos C2     = 8 sinA2.sinB2.⁡sin C2  
Solution: Here, 
         A + B + C = π^c
         A + B = π^c – C 
Taking sin and cos on both the sides then,
sin (A + B) =  sin (π^c-C) =  sin C   
cos (A+B) =  cos (π^c-C) =  - cos C
    
Taking,  sin 2A +sin⁡2B+  sin⁡2C,
           sin 2A +sin⁡2B+  sin⁡2C
         = 2⁡sin2A+2B2cos(2A-2B2)+ sin 2C
         = 2⁡sin A+Bcos(A-B) + 2sinC cos C
         = 2 sin C cosA-B+ 2sin C cos C
         = 2⁡sin C[cos (A-B)+ cos C ]
         = 2⁡sin C[cos (A-B) - cos(A-B)]
         = 2⁡sin C . 2sin A sin B⁡
         = 4 sin⁡ A sin⁡B sin⁡C.

Now, 
LHS = sin2A+sin2B+sin2C4⁡cos A2.⁡cos B2.cos C2      n
         = 4 sin A⁡sinB sinC4cos A2 cos B2 cos C2
        = 4 × 2 sin A2⁡⁡cos A2 × 2sin B2 ⁡cos B2 × 2⁡sinC2⁡cosC24⁡cos A2 ⁡cos B2 ⁡cos C2
       = 8 sinA2.sinB2.⁡sin C2 
       = RHS.

Proved

11. If A + B + C = π^c, prove that:

cos A + cos B+ cos C = 1 +4 sin A2.sinB2.sin⁡sin C2

Proof: Here, 
      A + B + C = π^c
 ⇒  A + B = π^c - C
⇒ 
A2+B2=πc2- C2
⇒  sin(A2+B2)= sin(πc2- C2)= cosC2
⇒  cos(A2+)= cos (πc2- C2)= sin C2

LHS = cos A + cos B + cos C
        = 2 cos A+B2.cos A-B2+ 1 - 2 sin² C2
       = 2 sin C2 cos A-B2- 2 sin² C2+ 1
       = 1 + 2 sin C2[cosA-B2- sin C2]
       = 1 + 2 cos C2[ cosA-B2- cos(A+B2)]
       = 1 + 2 sin C2.2sin12(A-B2+A+B2 ) Sin12(A+B2+A-B2
       = 1 + 4 sinC2.sin12A-B+A+B2. sin12A+B-A+B2
       = 1 + 4 sinC2.sinA2 SinB2
       = 1 +4 sin A2.sinB2.sin⁡sin C2
       = RHS.

Proved.

12. If A + B + C = 180^o, prove that: cotA cotB + cotB cotC + cotC cotA = 1.
Proof: Here, 
We have given,
         A + B + C = 180^o
 ⇒     A + B = 180 - C

Now operating on both sides by cot we get
         Cot(A + B) = cot(180 - C)
⇒       cotA cotB - 1cotA + cotB = - cotC
⇒      cotA cotB - 1 = -cotC( cotA + cotB)
⇒      cotA cotB - 1 = - cotC cotA - cotC cotB
⇒      cotA cotB + cotB cotC + cotC cotA = 1.
     LHS = RHS

Proved.

13. If A + B + C = 180^o, prove that:
tanA + tanB + tanC = tanA tanB tanC.
Proof: We have given,
     A + B + C = 180^o  
⇒     A + B = 180^o - C
Operating both sides by tan, we get
       tan(A + B) = tan( 180^o - C)
⇒     tanA +  cotB 1 - tanA tanB = - tanC
⇒      tanA + tanB = -tanC( 1 - tanA tanB)
⇒    tanA + tanB = - tanC + tanA tanB tanC
⇒     tanA + tanB + tanC = tanA tanB tanC
  ∴      LHS = RHS.

Proved.

14. If A + B + C = 180^o, prove that:
tan2A + tan2B + tan2C = tan2A tan2B tan2C.
Proof: We have given,
     A + B + C = 180^o  
⇒    2A + 2B = 180^o - 2C
Operating both sides by tan, we get
       tan(2A + 2B) = tan( 180^o - 2C)
⇒     tan2A+ tan2B 1 - tan2A tan2BcotA   = - tan2C
⇒     tan2A + tan2B = -tan2C( 1 - tan2A tan2B)
⇒     tan2A + tan2B = - tan2C + tan2A tan2B tan2C
⇒     tan2A + tan2B + tan2C = tan2A tan2B tan2C
       LHS = RHS

Proved.

15. If A , B & C are the vertices of a triangle, prove that:
cos2A - cos2B - cos2C = -1 + 4 cosA sinB sinC.
Proof: Here, we have given,
    A + B + C = 180^o ⟹ A + B = 180^o - C 

LHS = cos2A - cos2B - cos2C
        =  2sin2A + 2B2 sin2B - 2A2 - cos2C
       = 2sin(A + B) sin(B - A) - cos2C
       = 2sin(180^o - C) sin(B - A) - ( 1 - 2sin²C)
       = 2 sinC sin(B - A) - 1 + 2sin²C
       = - 1 + 2sinC[ sin(B - A) + sinC]
       = - 1 + 2sinC[ sin(B - A) + sin(180 - (A + B))]
       = - 1 + 2sinC[ sin(B - A) + sin(A + B)]  
       = - 1 + 2sinC ( 2 sinB cosA)
       = -1 + 4 cosA sinB sinC.
       = RHS.

Proved.

16. If A + B + C = π^c , prove that:
cos²A + cos²B + cos²C = 1 - 2cosA cosB cosC.
Proof: We have given,
     A + B + C = π^c 
⟹ A + B =  π^c - C
⟹  cos(A + B) = cos(π^c - C)
⟹   cos(A + B) = - cosC
⟹    cosA cosB - sinA sinB = - cosC
⟹   cosA cosB + cosC = sinA sinB
Squaring both sides,
      (cosA cosB + cosC)² = (sinA sinB)² 
⟹  cos²A cos²B + 2cosA cosB cosC + cos²C = ( 1 - cos²A)(1 - cos²B)     [ Since, sin²A = 1 - cos²A.]
⟹    cos²A cos²B + 2cosA cosB cosC + cos²C = 1 -    ⟹     cos²A- cos²B+ cos²A cos²B
⟹     cos2A +  cos²B + cos²C = 1 - 2cosA cosB cosC
  ∴    LHS = RHS.

Proved.

Note: Next method will be tought in class.
 
17. If A,B, C,D are cylic quadrilateral then prove that cosA+cosB+cosC+cosD=0.
Proof: We know from one of its many properties that, the opposite angles of ABCD are supplementary,
  That is    A + C = 180°……………….………………….(1)
       And    B + D = 180°……………….………………….(2)
Transposing C from left to right in eq. (1),
A = 180^o- C
Taking cosines on both sides,
          cos A = cos (π - C) = - cos C 
⟹    cos A + cos C = 0………………………..……(3)

Similarly it can be shown from eq.(2) that
          cos B + cos D = 0……………………..……………(4)

Adding eq.(3) and eq.(4), we get
  cos A + cos B + cos C + cos D = 0

Proved.

18. If A+B+C = 180° prove that : Sin2A + Sin2B + Sin2C = 4SinA SinB SinC.
Proof: Here,
LHS  = sin2A + sin 2B +sin2C
         = 2 sin(A+B)cos(A-B) + 2sinC cosC
         =2sinC cos(A-B)+2sinC cosC
         =2sinC (cos(A-b) + cos C)
         =2sin C(cos(A-B) - cos(A+B) )
         = 2sinC . 2sin A sin B
         =4 sinA sin B sin C
         = RHS.

Proved.

19. If A+B+C = 180^o, then prove that : sin A.cosB.cosC + sinB. cosA.cosC + sinC.cosA.cosB = sinA.sinA.sinC.
Proof: Here,
Given A+B+C =180^o.
Now,
 L.H.S. = sin A.cosB.cosC + sinB. cosA.cosC + sinC.cosA.cosB
            =cosC( sinA cosB+ sinB cosA) + sinC cosA cosB
            = cosC sin( A+B) + sinC cosA cosB
            = cosC sinC + sinC cosA cosB
            = sinC[ cosC + cosA cosB ]
            = sinC[ cosA cosB + cos{180 -(A+B) }]      [ Sincs, cosC= cos{180 - (A+B) } .]
            = sinC [ cosA cosB - cos( A+B) ]
            = sinC [ cosA cosB - cosA cosB+sinA sinB ]
            = sinA sinB sinC
            = R.H.S.

Proved.

N