Unit circle trig - have I got this right?

[Bahnanna]

Find exact value of tan x if cos x < sin x = -2/5, I got 2 over sq root 21

 

[Deleted member 4993]

Find exact value of tan x if cos x < sin x = -2/5, I got 2 over sq root 21

In your problem, is:

cos(x) = -2/5

or

sin(x) = -2/5

Very poorly expressed problem statement!!

 

[Bahnanna]

Unit circle trig

In the problem cos(x) is less than sin(x)=-2/5 so cos(x) is less than sin (x) and sin(x) is equal to -2/5, hope that helps

 

[Deleted member 4993]

In the problem cos(x) is less than sin(x)=-2/5 so cos(x) is less than sin (x) and sin(x) is equal to -2/5, hope that helps

Now find cos(x) using

cos²(x) = 1 - sin²(x)

 

[Bahnanna]

Unit circle trig

Thanks for the hint
I got -2 over the squared root of 21
Is that correct??

 

[Deleted member 4993]

Thanks for the hint
I got -2 over the squared root of 21
Is that correct??

No.

Please show your work.

 

[Bob Brown MSEE]

Find exact value of tan x if cos x < sin x = -2/5, I got 2 over sq root 21

c=-sqrt[1-ss] <= need to use neg sqrt because cos x < sin x
t=s/c=-2/5/-sqrt[1-ss]
t=2/(5sqrt[1-ss])
t=2/(5sqrt[1-4/25])
t=2/(5sqrt[21/25])

 

[Bahnanna]

Trig

so i did:

sin (x) = -2/5 (given)

cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (-2/5)^2
cos^2(x) = 21/25
cos(x) = root(21/25)
cos(x) = root(21) /5

sin(x)/cos(x) = (-2/5)/(root(21)/5)
tan(x) = (-2/5)*(5/root(21))
tan(x) = -2/root(21)

 

[Deleted member 4993]

so i did:

sin (x) = -2/5 (given)

cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (-2/5)^2
cos^2(x) = 21/25
cos(x) = root(21/25)
cos(x) = root(21) /5..... But your condition is that sin(x) > cos(x) . Is -2/5 greater than √(21)/5 ?

sin(x)/cos(x) = (-2/5)/(root(21)/5)
tan(x) = (-2/5)*(5/root(21))
tan(x) = -2/root(21)

Did you read the post by "Bob Brown"?