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Question Number 19352 by Tinkutara last updated on 10/Aug/17

Prove that ∣ z_{1} - z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}

- 2 ∣ z_{1} ∣ ∣ z_{2} ∣ \cos (θ_{1} - θ_{2})

Answered by ajfour last updated on 10/Aug/17

{(r_{1} \cos θ_{1} - r_{2} \cos θ_{2})}^{2} + {(r_{1} \sin θ_{1} - r_{2} \sin θ_{2})}^{2}

= r_{1}^{2} + r_{2}^{2} - {2 r}_{1} r_{2} (\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2})

=∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} - 2 ∣ z_{1} ∣∣ z_{2} ∣ \cos (θ_{1} - θ_{2}) .

Commented by Tinkutara last updated on 10/Aug/17

Thank you very much Sir!