Question

Prove that,
i) sin(A+B)=sinAcosB+cosAsinB​

Answer 1

Sin (A + B) = sinAcosB + cosAsinB

let, A=60' and B=30' ( here the sign ' bears degree)

L.H.S= sin (A + B)

=sin90' ( putting the values of A and B)

=1

R.H.S= sinAcosB + cosAsinB

= sin60'cos30' + cos60'sin30'

= sqrt3/2*sqrt3/2+ 1/2*1/2

=3/4+1/4

=4/4

=1  

so, L.H.S= R.H.S or sin(A + B) = sinAcosB + cosAsinB (proved)  

Geometrical method

Sin A+B=$\frac{PN}{OP}$

=$\frac{PR+RN}{OP}$

=$\frac{PR}{OP} +\frac{RN}{OP}$+$\frac{RN}{OM} *\frac{OM}{OP}$

=$\frac{PR}{PM} *\frac{PM}{OP}$

=Cos A*Sin B+SinA*CosB

=SinA+CosB+CosA SinB

=Sin(A-B)=SinA cosB-CosA SinB

Answer 2

Answer:

Kindly refer from Attachment, It is proven through Euler's Formula....