Question

Find the other 5 trig functions if cos=5/6

domain=3pi/2 11 years ago

Answer 1

Someone refresh me on this again..

Answer 2

ok so based on that domain
wer're on the 4th quadrant

meaning, cosine is positive, sine is negative, tangent is .. .negative becasue the other two differ in signs anyway

hypotenuse is radius, is always positive

Answer 3

Ohhh. I should solve for sin using a identity..

Answer 4

For a second I thought I was suppose to use the unit circle somehow..

Answer 5

sin^2+(5/6)^2=1

sin^2=sqrt11/6

Answer 6

well, not sin^2 but sin

Answer 7

then, i just use cos/sin for tangent, and inverse everything right?

Answer 8

then I just make sure only cos/sec are positive?

Answer 9

hmm actually lemme fix that too =)

Answer 10

hmmm wait a sec.. man l hold

Answer 11

I don't know, I remember solving these with identities. Maybe you can check my work?

cos=5/6
sin=-sqrt11/6
tan=-5(sqrt11)/11
sec=6/5
csc=-6(sqrt11)/11
cot=-sqrt11/5

Answer 12

\(\bf {\color{blue}{ cos}}(\theta)=\cfrac{adjacent}{hypotenuse}
\to \cfrac{5}{6}\to \cfrac{a=5}{c=6}
\\ \quad \\
c^2=a^2+{\color{blue}{ b}}^2\implies \pm\sqrt{c^2-a^2}={\color{blue}{ b}}\impliedby \textit{we know sine is negative}
\\ \quad \\
-\sqrt{c^2-a^2}={\color{blue}{ b}}\qquad sin(\theta)=\cfrac{{\color{blue}{ b}}}{c}\qquad tan(\theta)=\cfrac{{\color{blue}{ b}}}{a}\)

Answer 13

well... yes.. just that tangent is b/a a = 5

Answer 14

I got tangent wrong?

Answer 15

well... tangent \(\bf tan(\theta)=\cfrac{opposite}{adjacent}\)

Answer 16

I just put cos/sin,

5/6 times 6/radical 11

Answer 17

adjancent = 5
hypotenuse = 6
opposite = \(-\sqrt{11}\)

Answer 18

does that mean my sin and cos are also wrong

Answer 19

nope, your cosine and sine are correct,
but you have -> tan=-5(sqrt11)/11 <--

Answer 20

\(\bf {\color{brown}{ sin}}(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
{\color{blue}{ cos}}(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\

% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad

% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}

\\ \quad \\

% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}

\qquad \qquad

% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}\)

Answer 21

\[\frac{ 5 }{ 6 }\times -\frac{ 6 }{ \sqrt{11} }=\frac{ -30 }{ 6\sqrt{11} }=\frac{ -5 }{ \sqrt{11} }=\frac{ -5\sqrt{11} }{ 11 }\]
let's say we just used tan=cos/sin, what did I do wrong?

Answer 22

\(\bf tan(\theta)=\cfrac{opposite}{adjacent}\to \cfrac{\frac{-\sqrt{11}}{6}}{\frac{5}{6}}\to \cfrac{-\sqrt{11}}{\cancel{6}}\cdot \cfrac{\cancel{6}}{5}\)

Answer 23

lol you should of just told me tan=sin/cos, not cos/sin xD

Answer 24

hmmm well.... yes tan = sin /cos :)
"
cos=5/6 <-----
sin=-sqrt11/6 <-----
tan=-5(sqrt11)/11
sec=6/5
csc=-6(sqrt11)/11
cot=-sqrt11/5
"

Answer 25

yeah i see that now.

tan=-sqrt11/5
cot=-5sqrt11/11

basically a switch

Answer 26

yeap

Answer 27

cot and csc and sec
are just the other two upside-down

Answer 28

the other three rather

Answer 29

thank you :)

Answer 30

yw