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Oliga [24]
3 years ago
5

What are the coordinates of the terminal point corresponding to = –120? Use the Pythagorean theorem to prove that this point lie

s on the unit circle?
Mathematics
1 answer:
Vikentia [17]3 years ago
8 0
We know that
the unit circle has a radius equals 1 unit

-120°------> is an angle belong to the III quadrant
so
the x and y coordinates will be negative
x=r*cos ∅
y=r*sin ∅
where
r=1
and 
∅=180-120-----> ∅=60°

therefore
x=(1)*cos 60°-----> x=1/2-----> x=-1/2   (remember x is negative)
y=(1)*sin 60°-----> y=√3/2-----> y=-√3/2  (remember y is negative

<span>let's apply the Pythagorean theorem
</span>x²+y²-----> (-1/2)²+(-√3/2)²----> (1/4)+(3/4)-----> 1----> is ok because 1 is the radius of the unit circle

the answer is
the coordinates of the terminal point are (-1/2,-√3/2)


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Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x. (
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So it tells us that g(3) = -5 and g'(x) = x^2 + 7.

So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))

now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16 

Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84

So we have our linear approximation for the two. 

If you wanted to, you could check your answer by finding g(x).  Since you know g'(x), take the antiderivative and we will get 
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35

So just to check our linear approximations use that to find g(2.99) and g(3.01)

g(2.99) = -5.1597
g(3.01) = -4.8397

So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer.  Not a bad method if you ever need to use it. 
5 0
3 years ago
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