`-(13pi)/12` Find the exact values of the sine, cosine, and tangent of the angle.

Monday, April 23, 2012

$\displaystyle-\frac{{{13}\pi}}{{12}}$ Find the exact values of the sine, cosine, and tangent of the angle.

$\displaystyle{\sin{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}$

using the property sin(-x)=-sin(x),

$\displaystyle{\sin{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}=-{\sin{{\left(\frac{{{13}\pi}}{{12}}\right)}}}$

$\displaystyle=-{\sin{{\left(\frac{\pi}{{2}}+\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}$

Now using sin(pi/2+x)=cos(x),

$\displaystyle=-{\cos{{\left(\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}$

$\displaystyle=-{\left({\cos{{\left(\frac{\pi}{{3}}\right)}}}{\cos{{\left(\frac{\pi}{{4}}\right)}}}-{\sin{{\left(\frac{\pi}{{3}}\right)}}}{\sin{{\left(\frac{\pi}{{4}}\right)}}}\right)}$

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

$\displaystyle=\frac{{\sqrt{{{3}}}-{1}}}{{{2}\sqrt{{{2}}}}}$

rationalizing the denominator,

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

$\displaystyle{\cos{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}$

using the property cos(-x)=cos(x),

$\displaystyle{\cos{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}={\cos{{\left(\frac{{{13}\pi}}{{12}}\right)}}}$

$\displaystyle{\cos{{\left(\frac{{{13}\pi}}{{12}}\right)}}}={\cos{{\left(\frac{\pi}{{2}}+\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}$

now using cos(pi/2+x)=-sin(x),

$\displaystyle=-{\sin{{\left(\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}$

$\displaystyle=-{\left({\sin{{\left(\frac{\pi}{{3}}\right)}}}{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{\cos{{\left(\frac{\pi}{{3}}\right)}}}{\sin{{\left(\frac{\pi}{{4}}\right)}}}\right)}$

$\displaystyle=-{\left(\frac{\sqrt{{{3}}}}{{2}}\cdot\frac{{1}}{\sqrt{{{2}}}}+\frac{{1}}{{2}}\cdot\frac{{1}}{\sqrt{{{2}}}}\right)}$

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

rationalizing the denominator,

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

$\displaystyle{\tan{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}=\frac{{\sin{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}}{{\cos{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}}$

plug in the values of $\displaystyle{\sin{{\left(\frac{{-{13}\pi}}{{12}}\right)}}},{\cos{{\left(\frac{{-{13}\pi}}{{12}}\right)}}}$ obtained above,

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

$\displaystyle=-\frac{{\sqrt{{{3}}}-{1}}}{{\sqrt{{{3}}}+{1}}}$

rationalizing the denominator,

$\displaystyle=-\frac{{{\left(\sqrt{{{3}}}-{1}\right)}{\left(\sqrt{{{3}}}-{1}\right)}}}{{{\left(\sqrt{{{3}}}+{1}\right)}{\left(\sqrt{{{3}}}-{1}\right)}}}$