Question

Is it to right to say Sin (A+B)=
sin A+sinB ?Justify your answer.​

Answer 1

ANSWER:

No; sin(A + B)= sinA cosB + cosA sinB

Answer 2

sin(A+B)=sinA+sinB⇒sinAcosB+cosAsinB=sinA+sinB

⇒sinAcosB−sinA=sinB−cosAsinB⇒sinA(cosB−1)=sinB(1−cosA)⇒(cosB−1)sinB=(1−cosA)sinA⇒(cosB−1)2(sinB)2=(1−cosA)2(sinA)2[squaring with the fear of extraneous roots]

⇒(cosB−1)21−(cosB)2=(1−cosA)21−(cosA)2⇒(1−cosB)(1+cosB)=(1−cosA)(1+cosA)

[considering cosA and cosB are not equal to 1 and by dividendo and componendo we get ]

⇒cosA=cosB⇒A=2kπ±B

[k belongs to the set of integers]

Now putting A=2kπ± B to the original equation

sin(A+B)=sinA +sinB

>>sin(2kπ+-B +B)=sin(2kπ+-B) +sinB

gives

>>sin(2kπ+2B)=2sinB … …{1}

and

>>sin (2kπ) = -sinB+sinB =0 (an identity)… …{2}

from 1

>>sin2B =2sinB

>>2sinBcosB=2sinB

>>sinB=0 and cosB=1

But, cosB=1 will not be considered as mentioned previously.

We have got only solution i.e. sinB=0 implies

>> B= nπ (for all n belongs to integers) .

Now,

sin(A+B)=sinA +sinB

>>sin(A+nπ)=sinA +sinnπ

>>sinA =sinA

and it's also an identity.

So we can conclude thatsin(A+B)=sinA+sinB is false in general.

But if it is asked whensin(A+B)=sinA+sinBis true