Question

Prove the following results:
2cosacosb=cos(a+b)+cos(a-b)

Answer 1

Answer:

Cos(A+B) = Cos(A)*Cos(B) - Sin(A)*Sin(B)   --------(1)

Cos(A-B) = Cos(A)*Cos(B) + Sin(A)*Sin(B)  ---------(2)

Now your question

Cos(A+B) -Cos(A-B) =[Cos(A)*Cos(B) - Sin(A)*Sin(B)] - [ Cos(A)* Cos(B) + Sin(A)*Sin(B)]

                                  =Cos(A)*Cos(B) - Sin(A)*Sin(B) -  Cos(A)* Cos(B) - Sin(A)*Sin(B)

                                   = -2[Sin(A)*Sin(B)]

{ Since Cos(A)*Cos(B) -Cos(A)*Cos(B) =0}

So I guess it cant be proved or the question asked is wrong.

BTW you can also use the formula

Step-by-step explanation:

Answer 2

Answer:

Cos(A+B) = Cos(A)*Cos(B) - Sin(A)*Sin(B)   --------(1)

Cos(A-B) = Cos(A)*Cos(B) + Sin(A)*Sin(B)  ---------(2)

Now ↓↓

Cos(A+B) -Cos(A-B) =[Cos(A)*Cos(B) - Sin(A)*Sin(B)] - [ Cos(A)* Cos(B) + Sin(A)*Sin(B)]

                                  =Cos(A)*Cos(B) - Sin(A)*Sin(B) -  Cos(A)* Cos(B) - Sin(A)*Sin(B)

                                   =-2[Sin(A)*Sin(B)]{ Since Cos(A)*Cos(B) -Cos(A)*Cos(B) =0}

So I guess it cant be proved or the question asked is wrong.

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