MHB Precalculus help --> cos2x=3/5 and 90<x<180

[yeny]

Hello, can someone please help me with this problem?

I have to find the values of the 6 trig functions if the conditions provided hold

cos2(theta)=3/5

90 degrees is less than or equal to theta, and theta is also less than or equal to 180 degrees

THANK you so much.

 

[skeeter]

Hello, can someone please help me with this problem?

I have to find the values of the 6 trig functions if the conditions provided hold

cos2(theta)=3/5

90 degrees is less than or equal to theta, and theta is also less than or equal to 180 degrees

given $\cos(2\theta) = \dfrac{3}{5}$ and $\theta$ resides in quadrant II ...

$\cos(2\theta) = 2\cos^2{\theta} -1 = \dfrac{3}{5} \implies \cos^2{\theta} = \dfrac{4}{5} \implies \cos{\theta} = - \dfrac{2}{\sqrt{5}}$

$\sin^2{\theta} = \dfrac{1}{5} \implies \sin{\theta} = \dfrac{1}{\sqrt{5}}$

use your basic trig identities to determine the values of the remaining four