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1 year ago

For a particular angle Θ, the cosine function f(x) = a cos b(Θ) has the following values within one cycle of the function:

Mathematics

1 answer:

expeople1 [14]1 year ago

7 0

If f(θ) = a cos(bθ), then from the first condition we find

f(0) = 3 ⇒ a cos(0) = 3 ⇒ a = 3

Together with the other conditions, it's evident that f(θ) has a period of π, so

2π/b = π ⇒ b = 2

so that

f(θ) = 3 cos(2θ)

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Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

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$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

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${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

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We can choose any point for x and solve for y or vice-versa

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${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

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The point (5,0) and (-7,0) lies on the circle.

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