Proof for sin(A+B)=sinAcosB+cosAsinB

Can you guys provide me with the proof for sin(A+B)=sinAcosB+cosAsinB?? I have one but would like to know a few other methods as well.

Note by Prithwi Sinha
8 years ago

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5 votes

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Comments

Using Euler's formula, we can represent Sin and Cos in terms of the complex exponential:

eiθ =cosθ+isinθe^{i \theta}\ = \cos \theta + i\sin\theta

Since Sin is the imaginary part:

sinθ=Im(eiθ)\sin\theta = Im(e^{i\theta})

Applying to the question:

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

Expanding the exponential:

=Im(eiA×eiB)= Im(e^{iA}\times e^{iB})

Rewriting in terms of Sin and Cos:

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

Expanding the product:

=Im[cos(A)cos(B)+icos(A)sin(B)+isin(A)cos(B)+i2sin(A)sin(B)]= Im[\cos(A)\cos(B) + i\cos(A)\sin(B) + i\sin(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:

cosθ=Re(eiθ)\cos\theta = Re(e^{i\theta})

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!

Ryan Carson - 8 years ago

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Thanks Ryan!!But I guess this is quite above my wavelength.Can you please provide me with an easier one??Thanks for your time anyway!!

PRITHWI SINHA - 8 years ago

I know the common one draw a line ox on x axis (straight line)mark a point y above x and zoin oy let angle xoy be a , similarlly draw a point above y mark it z join oz and let angle yoz be b.
Next join a perpendicular to ox line mark it L. Then join zy in such a way that angle oyz is 90 degree (and a perpendicular to ox extended joining y then mark poin m thwm my=rl In triangle zoL sin (a+b)=zL/zo If we have a point R perpendicular to zL such that angle zRy=90degree Then zl/zo=zr/zo +rL/zo = zr/zo +my/zo =zr/zo × zy/zy +my/zo × oy/oy =zr/zy × zy/zo +my/oy × oy/zo zr/zy would be cos a as angle rzy Cos a×sin b+sin a as×cos b If unable to understand then search for a diagram

Tushar gautam - 8 years ago

This one is quite good for a simple geometrical understanding: http://demonstrations.wolfram.com/CosineAndSineOfTheSumOfTwoAngles/

Ben Blayney - 8 years ago

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Thanks Ben!!

PRITHWI SINHA - 8 years ago
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