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#### DreamWeaver

##### Well-known member

- Sep 16, 2013

- 337

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

prove that

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

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- Thread starter
- #1

- Sep 16, 2013

- 337

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

prove that

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

- Mar 31, 2013

- 1,347

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

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- Sep 16, 2013

- 337

Nicely done, Kaliprasad! Very efficient!