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

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To solve the problem of finding the six trigonometric functions given cos(2θ) = 3/5 and 90° < θ < 180°, the first step is to recognize that θ is in the second quadrant. From the identity cos(2θ) = 2cos²(θ) - 1, it follows that cos²(θ) = 4/5, leading to cos(θ) = -2/√5. Additionally, sin²(θ) = 1/5 gives sin(θ) = 1/√5. The remaining trigonometric functions can be derived using basic identities, completing the solution.
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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.
 
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yeny said:
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
 
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