Question

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

Answer 1

Refer to a solution using the Mathematica v9 computer program.

Answer 2

\( \huge cos(2θ)= \cos^2 \theta -\sin^2 \theta = \frac{1}{\sqrt{2}}\)
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 :-)

Answer 3

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