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NemiM [27]
3 years ago
13

Did ancient mathematicians use a ruler to determine the values of sine and cosine in a circle?

Mathematics
2 answers:
12345 [234]3 years ago
6 0

They didn’t use a ruler I think they made their own materials from what they had. Also thank you.

Fofino [41]3 years ago
5 0

Answer and Explanation:

No, they did not use a ruler to compute the values of trig functions. They used geometry to prove theorems about trigonometry, then used those results and ordinary computations involving addition, subtraction, multiplication, division, and square roots to determine the entries in trigonometric tables.

Claudius Ptolemy (85 C.E.–165) created the first trig tables that we know of, and he included proofs of the theorems he used. One of the main theorems he used is what is now called Ptolemy’s theorem.

In order to prove his sum and difference forumlas, Ptolemy first proved what we now call Ptolemy’s theorem.

<em>Ptolemy’s theorem</em>. The product of the diagonals of a cyclic quadrilateral is equal to the sum of the products of the opposite sides.

A cyclic quadrilateral is a quadrilateral inscribed in a circle as <em>ABCD</em>. Ptolemy’s theorem says:

<em>AC </em>⋅ <em>BD </em>= <em>AB </em>⋅ <em>CD </em>+ <em>AD </em>⋅ <em>BC</em>

<em />

When one of the diagonals (like ) is a diameter of the circle, that gives you a way to determine the trig functions for sum and differences of angles.

Sometimes, the sum and difference formulas for sine and cosine are still called Ptolemy’s formulas.

sin(<em>α</em> + <em>β</em>) = sin <em>α</em> cos <em>β</em> + cos <em>α</em> sin <em>β</em>

cos(<em>α</em> + <em>β</em>) = cos <em>α</em> cos <em>β</em> − sin <em>α</em> sin <em>β</em>

sin(<em>α</em> − <em>β</em>) = sin <em>α</em> cos <em>β</em> − cos <em>α</em> sin <em>β</em>

cos(<em>α</em> − <em>β</em>) = cos <em>α</em> cos <em>β</em> + sin <em>α</em> sin <em>β</em>

Ptolemy started with angles he knew the trig functions for. These included angles in various isosceles triangles: 45°-45°-90°, 60°-60°-60°, and 36°-72°-72°, the last one described by Euclid in Proposition IV.10. That gave him the trig functions for 36°, 60°, and 90°. Using half-angle formulas (mentioned by Hipparchus (190–120 B.C.E.), he could compute trig functions for half of those angles and a quarter of those angles. Then using the sum and difference formulas he computed the rest of the numbers in his trig table.

<em>Hope this helps!</em> :)

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Pachacha [2.7K]

Answer:

No

Step-by-step explanation:

The equation of a circle with center (a,b) and radius r is given as:

{(x - a)}^{2}  +  {(y - b)}^{2}  =  {r}^{2}

If a given point (x,y) does not lie on this circle, it will not satisfy its equation.

This means the distance from the point to the center is not equal to the radius.

It is either less or greater than the radius.

Hence you cannot write the equation of the circle.

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3 years ago
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Answer:

You did not include the price they bought the units at.

I will assume this price is $5.

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= $5,000

Recording it will be:

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7 0
2 years ago
Use substitution to solve the system of equations.<br> -1=2x-y<br> 8x-4y=-4
sineoko [7]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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UNO [17]

Answer:

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Step-by-step explanation:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

We have to find the roots of this given equation.

If a quadratic equation is of the form ax^{2}+bx +c=0

Its roots are \frac{-b+\sqrt{b^{2}-4ac } }{2a} and \frac{-b-\sqrt{b^{2}-4ac } }{2a}

Here the given equation is 2x^{2}-4x-1 = 0

a = 2

b = -4

c = -1

If the roots are x_{1} and x_{2}, then

x_{1} = \frac{-2+\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}

                       = \frac{4 +\sqrt{24}}{4}

                       = \frac{2+\sqrt{6} }{2}

x_{2} = \frac{-2-\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}

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6 0
3 years ago
1 1
andrey2020 [161]

Answer:

the answer should be 3/10

Step-by-step explanation:

if you look at the photo you can solve it from there

6 0
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