`csc(u - v)` Find the exact value of the trigonometric expression given that sin(u) = -7/25 and cos(v) = -4/5 (Both u and v are in quadrant III.)

Sunday, July 24, 2016

$\displaystyle{\csc{{\left({u}-{v}\right)}}}$ Find the exact value of the trigonometric expression given that sin(u) = -7/25 and cos(v) = -4/5 (Both u and v are in quadrant III.)

Given $\displaystyle{\sin{{\left({u}\right)}}}=-\frac{{7}}{{25}},{\cos{{\left({v}\right)}}}=-\frac{{4}}{{5}}$

using pythagorean identity,

$\displaystyle{{\sin}}^{{2}}{\left({u}\right)}+{{\cos}}^{{2}}{\left({u}\right)}={1}$

$\displaystyle{{\left(-\frac{{7}}{{25}}\right)}}^{{2}}+{{\cos}}^{{2}}{\left({u}\right)}={1}$

$\displaystyle{{\cos}}^{{2}}{\left({u}\right)}={1}-\frac{{49}}{{625}}$

$\displaystyle{{\cos}}^{{2}}{\left({u}\right)}=\frac{{{625}-{49}}}{{625}}=\frac{{576}}{{625}}$

$\displaystyle{\cos{{\left({u}\right)}}}=\sqrt{{\frac{{576}}{{625}}}}$

$\displaystyle{\cos{{\left({u}\right)}}}=\pm\frac{{24}}{{25}}$

since u is in Quadrant III ,

$\displaystyle{\cos{{\left({u}\right)}}}=-\frac{{24}}{{25}}$

Now $\displaystyle{{\sin}}^{{2}}{\left({v}\right)}+{{\cos}}^{{2}}{\left({v}\right)}={1}$

$\displaystyle{{\sin}}^{{2}}{\left({v}\right)}+{{\left(-\frac{{4}}{{5}}\right)}}^{{2}}={1}$

$\displaystyle{{\sin}}^{{2}}{\left({v}\right)}={1}-\frac{{16}}{{25}}=\frac{{9}}{{25}}$

$\displaystyle{\sin{{\left({v}\right)}}}=\sqrt{{\frac{{9}}{{25}}}}$

$\displaystyle{\sin{{\left({v}\right)}}}=\pm\frac{{3}}{{5}}$

since v is in Quadrant III ,

$\displaystyle\therefore{\sin{{\left({v}\right)}}}=-\frac{{3}}{{5}}$

$\displaystyle{\csc{{\left({u}-{v}\right)}}}=\frac{{1}}{{{\sin{{\left({u}-{v}\right)}}}}}$

$\displaystyle=\frac{{1}}{{{\sin{{\left({u}\right)}}}{\cos{{\left({v}\right)}}}-{\cos{{\left({u}\right)}}}{\sin{{\left({v}\right)}}}}}$

$\displaystyle=\frac{{1}}{{{\left(-\frac{{7}}{{25}}\right)}{\left(-\frac{{4}}{{5}}\right)}-{\left(-\frac{{24}}{{25}}\right)}{\left(-\frac{{3}}{{5}}\right)}}}$

$\displaystyle=\frac{{1}}{{\frac{{28}}{{125}}-\frac{{72}}{{125}}}}$

$\displaystyle=-\frac{{125}}{{44}}$

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