Auxiliary angle proof

[DreamWeaver]

By considering the product of complex numbers


\(\displaystyle (a+ib)\, (\cos \theta +i\sin \theta)\)


prove that


\(\displaystyle b\cos \theta+a\sin\theta=\sqrt{a^2+b^2}\, \sin \left(\theta+\tan^{-1}\frac{b}{a}\right)\)

 

[kaliprasad]

Spoiler

let a+ ib = r cos t + i sin t ..1

So a = r cos t
b = r sin t

Square and add to get r = (a^2+b^2)
divide to get t = arctan(b/a) ..2

So a + ib = (a^2+b^2) ^(1/2) ( cos t + i sin t)

Now (a+ib) (cos θ + i sin θ) = ( acos θ - b sin θ)+ i (a sin θ + bcos θ)

Firther (a^2+b^2) ^(1/2) ( cos t + i sin t) (cos θ + i sin θ)

= (a^2+b^2) ^(1/2) (cos (t+θ) + i sin (t+θ)

Comparing imaginary parts of both sides

(a sin θ + bcos θ) = (a^2+b^2) ^(1/2) sin (t+θ)
= (a^2+b^2) ^(1/2) sin (arctan (b/a) +θ)
Proved

 

[DreamWeaver]

Nicely done, Kaliprasad! Very efficient!