Is the Calculation for Sin and Tan Correct Given Cos in the Third Quadrant?

[andrewkg]

Q
If cos(alpha)=-sqrt3/5 and alpha is in the third quadrant, find exact values for sin(alpha) and tan(alpha).

A. Well what I did was use the pythagorean theorem so (-sqrt3)^2+(opposite)^2=5^2. then in the end I got sqrt22 for opposite so then sin(alpha)=sqrt22/5 and tan alpha =sqrt22/-sqrt3,
but here is the problem then the alpha is outside the third quadrant. then no matter the switchs made to my answs for sin tan +- they angles never fit properly. am I just making a dumb mistake of is the question off?

 

[Simon Bridge]

You should sketch the situation.
The opposite said should be -√22 (spot the minus sign). Pythagoras will only tell you the lengths, not the directions.

Try it this way:
Construct the right-angled triangle in the third quadrant. The angle to the hypotenuse, measured anticlockwise from the x-axis is $\alpha$ but you want to use trigonometry on interior angles.
The interior angle is $\theta = \alpha-180$ $H=5$ and $A=\sqrt{3}$... all positive values.
So $\alpha = 180+\arccos(\sqrt{3}/5)$ will be in the third quadrant.

 

[andrewkg]

Ok I see so sin(alpha)=-sqrt22/5 and tan alpha =-sqrt22/-sqrt3,

 

[Simon Bridge]

Care is needed:
tan(θ)=1 could come from O/A = 1/1 or O/A -1/-1 ... are they the same angle?
... a calculator evaluating tan^-1(1) will say 45deg for both.

 

[andrewkg]

Ok I need to come back to this its almost 2 am where I live and I cannot think right now. Thanks A ton for the help, my brain is just not functioning

 

[Simon Bridge]

2am effect - know it well - sweet dreams.

 

[HallsofIvy]

It would help a lot if you used parentheses. Do you mean sin(alpha)= -sqrt(3)/5 or -sqrt(3/5)?

 

[andrewkg]

Its sqrt(3)/5
Ok so could I just ust sin(alpha)=sin(pi/2+arccos(sqrt(3)/5)=-sqrt(22)/5
And tan (alpha)=the answer above ^/cos (pi/2+arccos(sqrt(3)/5)=-sqrt(22)/-sqrt3.

 

[andrewkg]

Because by doing this I have given an exact angle in the third quadrent correct.

 

[andrewkg]

Also not pi/2 just pi

 

[SammyS]

Its sqrt(3)/5
Ok so could I just use sin(alpha)=sin(pi/2+arccos(sqrt(3)/5)=-sqrt(22)/5
And tan (alpha)=the answer above ^/cos (pi/2+arccos(sqrt(3)/5)=-sqrt(22)/-sqrt3.

Because by doing this I have given an exact angle in the third quadrent correct.

Also not pi/2 just pi

(You can use the Edit feature to edit a previous post, rather than tacking on sentence fragments in additional posts.)


Sure. If $\ \alpha\$ is in the third quadrant and $\displaystyle \ \cos(\alpha)=-\frac{\sqrt{5}}{3}\,,\$ then $\displaystyle \ \alpha\ =\pi+\arccos\left(\frac{\sqrt{5}}{3}\right)\ .$

Then $\displaystyle \ \sin(\alpha)=\sin\left(\pi+\arccos\left(\frac{\sqrt{5}}{3}\right)\right)$

$\displaystyle \quad\quad\ \quad\quad\ \quad =\sin(\pi)\cdot\cos\left(\arccos\left(\frac{\sqrt{5}}{3}\right)\right)<br /> +\cos(\pi)\cdot\sin\left(\arccos\left(\frac{\sqrt{5}}{3}\right)\right)$

...​

 

[epenguin]

Are you sure you read and wrote the question right?

cos α = -√3/5 or cos α = -√(3/5) ?

 

[andrewkg]

I'm positive. cos(a)=-sqrt(3)/5

 

[cepheid]

You know, there are ways to solve this that are a lot easier. For instance, I found the following trig identities to be very useful:

sin²(α) + cos²(α) = 1

tan(α) = sin(α)/cos(α)

 

[andrewkg]

Ok so now I see the easier solution. sin a = -sqrt(1-(-sqrt(3)/5)) and tan a =-(sin a)/-(cos a)
Thanks cepheid and everyone else