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

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

I meant cosθ=2/3. Sorry. :(

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

|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 :)

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 \[\tan(\theta) = \frac{\sqrt{5}}{2}\]

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

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

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

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