Question

Let theta be an angle in quadrant 3 such that
cos theta = -1/4 .

Find the exact values of csc theta and cot theta.

Answer 1

\(\bf cos(\theta)=-\cfrac{1}{4} \implies \cfrac{\textit{opposite side}}{\textit{hypotenuse}} \implies \cfrac{b}{c}\\
c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2} = a\)

the angle is in Quadrant III
thus at III the "x" and "y" are both negative, and consequently cosine and sine are both negative, thus the value from the pythagorean theorem, will be negative for the side "a"

once you have "a", "b", and "c"
you can use that to find other trig identities :)

Answer 2

ark... shoot one sec, that's a bit wrong :(

Answer 3

\(\bf cos(\theta)=-\cfrac{1}{4} \implies \cfrac{\textit{adjacent side}}{\textit{hypotenuse}} \implies \cfrac{a}{c}\\
c^2 = a^2 + b^2 \implies \sqrt{c^2-a^2} = b\)

thus you'd need side "b" rather, is the only side missing in the pythagorean theorem

Answer 4

just min i will try to solve it

Answer 5

so a = -sqrt(17)

Answer 6

\(\bf c^2 = a^2 + b^2 \implies \sqrt{c^2-a^2} = b \implies
\sqrt{(4)^2-(-1)^2} \implies \sqrt{16-1}\)

Answer 7

so

csc theta = -4/sqrt(15)

cot theat = -1/sqrt(15)

right?

Answer 8

yes

Answer 9

hmmm

Answer 10

\(\bf a = -1 \qquad b = -\sqrt{15} \qquad c = 4\\
csc(\theta) = \cfrac{c}{b} \implies \cfrac{4}{-\sqrt{15}}\\
cot(\theta) = \cfrac{a}{b} \implies \cfrac{-1}{-\sqrt{15}}\)

Answer 11

the cotangent function will yield positive, the values are otherwise correct