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Mathematics 7 Online
OpenStudy (anonymous):

Find the exact value of the trigonometric function. Do not use a calculator. cot(-5pi/4)

OpenStudy (anonymous):

@ phi

OpenStudy (anonymous):

@freckles

OpenStudy (freckles):

do you know that cotangent is an odd function

OpenStudy (anonymous):

i did not

OpenStudy (freckles):

that means cot(-x)=-cot(x)

OpenStudy (anonymous):

so confusing lol

OpenStudy (freckles):

\[\cot(\frac{-5\pi}{4})=-\cot(\frac{5\pi}{4})\]

OpenStudy (freckles):

sin, csc, tan, cot are all odd functions cos, sec are even functions

OpenStudy (freckles):

anyways you should see 5pi/4 on the unit circle

OpenStudy (freckles):

@Tazmaniadevil do you see 5pi/4 on the unit circle should be in quadrant 3

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

umm i do not

OpenStudy (freckles):

do you see 225 deg because that is the same thing in degrees

OpenStudy (anonymous):

yes i think

OpenStudy (freckles):

ok what is the x and y coordinate given there

OpenStudy (anonymous):

wheres the cordinates

OpenStudy (freckles):

the coordinates are the things in ( , )

OpenStudy (freckles):

(x,y) x is the cosine value y is the sine value

OpenStudy (anonymous):

so it was the one with the twos right on the left side

OpenStudy (freckles):

both x and y are -sqrt(2)/2 there \[\cot(-\frac{5\pi}{4})=-\cot(\frac{5\pi}{4})=- \frac{\cos(\frac{5\pi}{4})}{\sin(\frac{5\pi}{4})}=-\frac{\frac{-\sqrt{2}}{2}}{\frac{-\sqrt{2}}{2}}\]

OpenStudy (freckles):

you can simplify that

OpenStudy (anonymous):

smh how do i simplify that ?

OpenStudy (freckles):

what is 5/5 =? 6/6=? 3/3=?

OpenStudy (anonymous):

1

OpenStudy (freckles):

12/12= -12/-12=?

OpenStudy (freckles):

yes

OpenStudy (freckles):

so you have -1 because there is negative in front of that mess

OpenStudy (anonymous):

right

OpenStudy (anonymous):

@phi

OpenStudy (anonymous):

@dan815

OpenStudy (freckles):

@tazmaniadevil did you have another question

OpenStudy (anonymous):

yes is -1 my answer

Nnesha (nnesha):

\(\color{blue}{\text{Originally Posted by}}\) @freckles so you have -1 because there is negative in front of that mess \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @Tazmaniadevil yes is -1 my answer \(\color{blue}{\text{End of Quote}}\)

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