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# 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
4 years, 2 months ago

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Using 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! · 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!! · 4 years, 2 months ago

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

Thanks Ben!! · 4 years, 1 month ago