Choose the point on the terminal side of -45°. (-3, -3) (4, -4) (5, 5) (-2, 2)

2 answers:

8 0

The angle of -45 its terminal side lies in the fourth quadrant, i.e. the x-coordinates is *positive* and y-coordinates is *negative*

The point (4,-4) is the only one that satisfies that condition

hope that helps

Sladkaya [172]2 years ago

4 0

(4, -4) is the answer

-45° is in quadrant IV and the point (4, -4) is the only point out of the points given that is also in quadrant IV

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Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

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$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

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