Let θ be an angle such that cosθ=23 and sinθ

Question

Let θ be an angle such that cosθ=23 and sinθ<0. Find tanθ

anonymous · Answer

Cosine cannot be larger than 1. Its range is between -1 and 1. Arc-cosine of 23 = NaN

anonymous · Answer

I meant cosθ=2/3. Sorry. :(

anonymous · Answer

If I'm not mistaken, theta is 48,189.. and therefore tan(theta) is 1,11..

mathsolver · Answer

|dw:1363106210176:dw|If you draw the cos curve ( as i have very badly) you will see that arccos(2/3) has two solutions, one at 48.19 and one at 311.81. since sin48.19 is above zero our thetha must be 311.81. so tan311.81=.1.118 and that is our answer :)

anonymous · Answer

sub your value of cos into

$$\cos^2 \theta + \sin^2 \theta = 1$$

to get
$$\sin^2 \theta  = \frac{5}{9}$$

as sin(theta) < 0 we know now that

$$\sin(\theta) = \frac{\sqrt{5}}{3}$$


now tan = sin/cos gives

$$\tan(\theta) = \frac{\sqrt{5}}{2}$$

anonymous · Answer

sorry that's meant to be $$\sin(\theta) = -\sqrt5/3$$

and $$\tan \theta =  - \sqrt5/2$$

raden · Answer

if θ in the quadrant 4th, then tan must be negative

anonymous · Answer

Thank you for the help everyone! Yeah I had $$\sqrt{5}/2$$ but I had forgot the negative.