## anonymous one year ago Find the value for cos(θ) if the following conditions hold: cos(2θ)=1/root2 and 180degrees<θ<270degrees

1. anonymous

Refer to a solution using the Mathematica v9 computer program.

2. anonymous

$$\huge cos(2θ)= \cos^2 \theta -\sin^2 \theta = \frac{1}{\sqrt{2}}$$ Rationalize $$\huge \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\huge \cos^2 \theta -\sin^2 \theta = \frac{\sqrt{2}}{2}$$ $$\huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\sin^2 \theta$$ Now we use this identity $$\huge \sin^2 \theta = \frac{1-\cos(2\theta)}{2}$$ $$\huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\frac{1-cos(2\theta)}{2}$$ Remember $$\huge \cos(2\theta) = \frac{\sqrt{2}}{2}$$ $$\huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\frac{1-\frac{\sqrt{2}}{2}}{2}$$\) $$\huge \cos^2 \theta = \frac{2+\sqrt{2}}{4}$$ Now square each side $$\huge \sqrt{\cos^2 \theta} = \sqrt{\frac{2+\sqrt{2}}{4}}$$ $$\huge \cos \theta = \frac{\sqrt{2+\sqrt{2}}}{2}$$ And since we are in the third quadrant it will be $$\huge \cos \theta = -\frac{\sqrt{2+\sqrt{2}}}{2}$$ WOW that was a lot of work :-)

3. anonymous

You're taking cosine of 2Θ so multiply the interval by 2 $$\cos 2\theta = \frac{ 1 }{ \sqrt2 }=\frac{ \sqrt2 }{ 2 }$$ , 360°<Θ<540° $2\theta=405°$ $\theta=202.5°$