Given-a-2-b-2-1-and-c-2-d-2-1-The-minimum-value-of-ac-bd-2-is-

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

Given a^{2} + b^{2} = 1 and c^{2} + d^{2} = 1

The minimum value of a c + b d - 2 is \dots

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 (a c + b d) ⩾ 0

2 + 2 (a c + b d) ⩾ 0

(a c + b d) ⩾ - 1

Commented by Joel578 last updated on 31/Dec/17

t h a n k y o u v e r y m u c h

Answered by ajfour last updated on 31/Dec/17

l e t a = \cos θ, b = \sin θ

c = \cos ϕ, d = \sin ϕ

a c + b d - 2 = \cos (θ - ϕ) - 2

m i n . v a l u e = - 3 .

Commented by Joel578 last updated on 02/Jan/18

t h a n k y o u v e r y m u c h

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