`tan u = 3/5, 0

Wednesday, December 3, 2008

$\displaystyle{\tan{{u}}}=\frac{{3}}{{5}},{0}$

Given $\displaystyle{\tan{{\left({u}\right)}}}=\frac{{3}}{{5}},{0}<{u}<\frac{\pi}{{2}}$

Angle u is in quadrant 1. A right triangle can be drawn is quadrant 1. Since

$\displaystyle{\tan{{\left({u}\right)}}}=\frac{{3}}{{5}}$ , the side opposite angle u is 3 and the side adjacent to angle u is 5. Using the pythagorean theorem the hypotenuse of the triangle is $\displaystyle\sqrt{{34}}.$

$\displaystyle{\sin{{\left({2}{u}\right)}}}={2}{\sin{{\left({u}\right)}}}{\cos{{\left({u}\right)}}}={2}{\left(\frac{{3}}{\sqrt{{34}}}\right)}{\left(\frac{{5}}{\sqrt{{34}}}\right)}=\frac{{30}}{{34}}=\frac{{15}}{{17}}$

$\displaystyle{\cos{{\left({2}{u}\right)}}}={1}-{2}{{\sin}}^{{2}}{\left({u}\right)}={1}-{2}{{\left(\frac{{3}}{\sqrt{{34}}}\right)}}^{{2}}={1}-\frac{{18}}{{34}}=\frac{{34}}{{34}}-\frac{{18}}{{34}}=\frac{{16}}{{34}}=\frac{{8}}{{17}}$

$\displaystyle{\tan{{\left({2}{u}\right)}}}=\frac{{{2}{\tan{{\left({u}\right)}}}}}{{{1}-{{\tan}}^{{2}}{\left({u}\right)}}}=\frac{{{2}{\left(\frac{{3}}{{5}}\right)}}}{{{1}-{{\left(\frac{{3}}{{5}}\right)}}^{{2}}}}=\frac{{\frac{{6}}{{5}}}}{{{1}-\frac{{9}}{{25}}}}=\frac{{\frac{{6}}{{5}}}}{{\frac{{25}}{{25}}-\frac{{9}}{{25}}}}=\frac{{\frac{{6}}{{5}}}}{{\frac{{16}}{{25}}}}=\frac{{15}}{{8}}$

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Wednesday, December 3, 2008

$\displaystyle{\tan{{u}}}=\frac{{3}}{{5}},{0}$

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