Question

prove sin(A + B)= sinAcosB+cosAsinB​

Answer 1

Step-by-step explanation:

θ

)

Applying to the question:

\sin(A+B) = Im(e^{i(A+B)})sin(A+B)=Im(e

i(A+B)

)

Expanding the exponential:

= Im(e^{iA}\times e^{iB})=Im(e

iA

×e

iB

)

Rewriting in terms of Sin and Cos:

= Im[cos(A) + i\sin(A)) \times (\cos(B) + i\sin(B)) ]=Im[cos(A)+isin(A))×(cos(B)+isin(B))]

Expanding the product:

= Im[\cos(A)\cos(B) + i\cos(A)\sin(B) + i\sin(A)\cos(B) + i^2\sin(A)\sin(B)]=Im[cos(A)cos(B)+icos(A)sin(B)+isin(A)cos(B)+i

2

sin(A)sin(B)]

Collecting into the real and imaginary parts:

= Im[(\cos(A)\cos(B) - \sin(A)\sin(B)) + i(\cos(A)\sin(B) + \sin(A)\cos(B))=Im[(cos(A)cos(B)−sin(A)sin(B))+i(cos(A)sin(B)+sin(A)cos(B))

Therefore:

\sin(A+B) = \cos(A)\sin(B) + \sin(A)\cos(B)sin(A+B)=cos(A)sin(B)+sin(A)cos(B)

as required!

(This method also makes it really simple to find \cos(A+B)cos(A+B), as we can see from Euler's formula:

\cos\theta = Re(e^{i\theta})cosθ=Re(e

iθ

)

Therefore: \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)cos(A+B)=cos(A)cos(B)−sin(A)sin(B)

Hope this helps!

Answer 2

Answer:

$hey$

Step-by-step explanation:

Here are two for the price of one (using Euler's formula):

cos(A+B)+isin(A+B)≡ei(A+B)≡eiA×eiB

cosABisinABeiABeiAeiB

≡[cos(A)+isin(A)][cos(B)+isin(B)]

cosAisinAcosBisinB

≡[cos(A)cos(B)−sin(A)sin(B)]+i[sin(A)cos(B)+cos(A)sin(B)]

cosAcosBsinAsinBisinAcosBcosAsinB

Now equate imaginary parts to give the result for sin(A+B)sinAB (and, if you want, equate real parts to give the result for cos(A+B)cosAB).