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.

No vote yet

5 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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

\(e^{i \theta}\ = \cos \theta + i\sin\theta\)

Since Sin is the imaginary part:

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

Applying to the question:

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

Expanding the exponential:

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

Rewriting in terms of Sin and Cos:

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

Expanding the product:

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

Collecting into the real and imaginary parts:

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

Therefore:

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

as required!

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

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

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

Hope this helps! – Ryan Carson · 4 years, 2 months ago

Log in to reply

– Prithwi Sinha · 4 years, 2 months ago

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!!Log in to reply

This one is quite good for a simple geometrical understanding: http://demonstrations.wolfram.com/CosineAndSineOfTheSumOfTwoAngles/ – Ben Blayney · 4 years, 2 months ago

Log in to reply

– Prithwi Sinha · 4 years, 1 month ago

Thanks Ben!!Log in to reply

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 · 4 years, 2 months ago

Log in to reply