How do you find sin, cos, tan, sec, csc, and cot given (-4,-4)?

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Answer 1 · 2018-03-02T08:49:28

See below.

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If we are given coordinates of the form $(x,y)$ , where $x$ and $y$ are negative then we are in the III quadrant.

Since $(-4,-4)$ are the sides of a right triangle, then the length of the terminal side ( the hypotenuse ) is given by Pythagoras' theorem:

Let the terminal side be $bbr$

$r^2=(-4)^2+(-4)^2$

$=>r=sqrt((-4)^2+(-4)^2)=4sqrt(2)$

So for the right triangle $bb(ABC)$ , we have:

$c=4sqrt(2)$

$a=-4$

$b=-4$

$sin(theta)="opposite"/"hypotenuse"=a/c=-4/(4sqrt(2))=color(blue)(-sqrt(2)/2)$

$cos(theta)="adjacent"/"hypotenuse"=b/c=-4/(4sqrt(2))=color(blue)(-sqrt(2)/2)$

$tan(theta)="opposite"/"adjacent"=a/b=(-4)/-4=color(blue)(1)$

Since:

$color(red)bb(csc(theta)=1/sin(theta))$

$color(red)bb(sec(theta)=1/cos(theta))$

$color(red)bb(cot(theta)=1/tan(theta))$

We have:

$csc(theta)=1/(-sqrt(2)/2)=color(blue)(-sqrt(2))$

$sec(theta)=1/(-sqrt(2)/2)=color(blue)(-sqrt(2))$

$cot(theta)=1/1=color(blue)(1)$