Asked by coco
                1) given cotθ = 1/2√7, find sinθ and cosθ in quadrant I
2) given tanθ = √5, find secθ and cotθ in quadrant III
            
        2) given tanθ = √5, find secθ and cotθ in quadrant III
Answers
                    Answered by
            Steve
            
    recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
r^2 = x^2+y^2
So, in QIII, where x and y are both negative,
tanθ = √5 = √5/1, so
y = -√5
x = -1
r = √6
cotθ = x/y = 1/√5
secθ = r/x = -√6
For #1, do you mean (1/2) √7 or 1/(2√7)?
In any case, follow the steps above.
    
sinθ = y/r
cosθ = x/r
tanθ = y/x
r^2 = x^2+y^2
So, in QIII, where x and y are both negative,
tanθ = √5 = √5/1, so
y = -√5
x = -1
r = √6
cotθ = x/y = 1/√5
secθ = r/x = -√6
For #1, do you mean (1/2) √7 or 1/(2√7)?
In any case, follow the steps above.
                    Answered by
            coco
            
    for #1 i mean (1/2) √7
which 1/2 be the x and √7 be the y?
    
which 1/2 be the x and √7 be the y?
                    Answered by
            coco
            
    would*
    
                    Answered by
            Steve
            
    so, cotθ = (√7)/2, meaning
x = √7
y = 2
since cotθ = x/y
    
x = √7
y = 2
since cotθ = x/y
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.