Notes: `sec(u) = -2, pi |

Friday, 30 May 2014

$\displaystyle{\sec{{\left({u}\right)}}}=-{2},\pi{\mid}$

Given $\displaystyle{\sec{{\left({u}\right)}}}=-{2},\pi<{u}<\frac{{{3}\pi}}{{2}}$

Angle u is in quadrant 3. If $\displaystyle{\sec{{\left({u}\right)}}}=-{2}$ then $\displaystyle{\cos{{\left({u}\right)}}}=-\frac{{1}}{{2}}.$

If a right triangle is drawn in quadrant 3, the side adjacent to angle u would be 1 and the hypotenuse of the triangle would be 2. Using the pythagorean theorem the third side would be $\displaystyle\sqrt{{{3}}}.$

$\displaystyle{\sin{{\left({2}{u}\right)}}}={2}{\sin{{\left({u}\right)}}}{\cos{{\left({u}\right)}}}={2}{\left(-\frac{\sqrt{{3}}}{{2}}\right)}{\left(-\frac{{1}}{{2}}\right)}=\frac{{{2}\sqrt{{3}}}}{{4}}=\frac{\sqrt{{3}}}{{2}}$

$\displaystyle{\cos{{\left({2}{u}\right)}}}={2}{{\cos}}^{{2}}{u}-{1}={2}{{\left(-\frac{{1}}{{2}}\right)}}^{{2}}-{1}={2}{\left(\frac{{1}}{{4}}\right)}-{1}=-\frac{{1}}{{2}}$

$\displaystyle{\tan{{\left({2}{u}\right)}}}=\frac{{{2}{\tan{{\left({u}\right)}}}}}{{{1}-{{\tan}}^{{2}}{\left({u}\right)}}}=\frac{{\frac{{{2}\sqrt{{3}}}}{{1}}}}{{{1}-{{\left(\frac{\sqrt{{3}}}{{1}}\right)}}^{{2}}}}=\frac{{{2}\sqrt{{3}}}}{{{1}-{3}}}=\frac{{{2}\sqrt{{3}}}}{{-{{2}}}}=-\sqrt{{3}}$

Given $\displaystyle{\sec{{\left({u}\right)}}}=-{2},\pi<{u}<\frac{{{3}\pi}}{{2}}$

Angle u is in quadrant 3. If $\displaystyle{\sec{{\left({u}\right)}}}=-{2}$ then $\displaystyle{\cos{{\left({u}\right)}}}=-\frac{{1}}{{2}}.$

$\displaystyle{\cos{{\left({2}{u}\right)}}}={2}{{\cos}}^{{2}}{u}-{1}={2}{{\left(-\frac{{1}}{{2}}\right)}}^{{2}}-{1}={2}{\left(\frac{{1}}{{4}}\right)}-{1}=-\frac{{1}}{{2}}$

Notes

Friday, 30 May 2014

$\displaystyle{\sec{{\left({u}\right)}}}=-{2},\pi{\mid}$

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Is there any personification in "The Tell-Tale Heart"?

Friday, 30 May 2014

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Is there any personification in &quot;The Tell-Tale Heart&quot;?

Is there any personification in "The Tell-Tale Heart"?