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Question Number 26917 by Joel578 last updated on 31/Dec/17

$$\mathrm{Given}{a}^2+b^2=1\mathrm{and}{c}^2+d^2=1$$$$\mathrm{The}\mathrm{minimum}\mathrm{value}\mathrm{of}ac+bd-2\mathrm{is}...$$

Commented by prakash jain last updated on 31/Dec/17

$${(a+c)}^2+{(b+d)}^2⩾0$$$$a^2+c^2+b^2+d^2+2(ac+bd)⩾0$$$$2+2(ac+bd)⩾0$$$$(ac+bd)⩾-1$$

Commented by Joel578 last updated on 31/Dec/17

$$thankyouverymuch$$

Answered by ajfour last updated on 31/Dec/17

$$leta=\cos \theta ,b=\sin \theta$$$$c=\cos \varphi ,d=\sin \varphi$$$$ac+bd-2=\cos (\theta -\varphi )-2$$$$min.value=-3.$$

Commented by Joel578 last updated on 02/Jan/18

$$thankyouverymuch$$

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