`-(7pi)/12` Find the exact values of the sine, cosine, and tangent of the angle.

Sunday, May 11, 2008

$\displaystyle-\frac{{{7}\pi}}{{12}}$ Find the exact values of the sine, cosine, and tangent of the angle.

$\displaystyle{\sin{{\left(\frac{{-{7}\pi}}{{12}}\right)}}}=-{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}$

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

using the identity $\displaystyle{\sin{{\left({x}+{y}\right)}}}={\sin{{\left({x}\right)}}}{\cos{{\left({y}\right)}}}+{\cos{{\left({x}\right)}}}{\sin{{\left({y}\right)}}}$

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

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

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

rationalizing the denominator,

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

$\displaystyle{\cos{{\left(\frac{{-{7}\pi}}{{12}}\right)}}}={\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}$

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

using the identity $\displaystyle{\cos{{\left({x}+{y}\right)}}}={\cos{{\left({x}\right)}}}{\cos{{\left({y}\right)}}}-{\sin{{\left({x}\right)}}}{\sin{{\left({y}\right)}}}$

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

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

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

rationalizing the denominator,

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

$\displaystyle=\frac{{\sqrt{{{2}}}-\sqrt{{{6}}}}}{{4}}$

$\displaystyle{\tan{{\left(\frac{{-{7}\pi}}{{12}}\right)}}}$

$\displaystyle=\frac{{\sin{{\left(\frac{{-{7}\pi}}{{12}}\right)}}}}{{\cos{{\left(\frac{{-{7}\pi}}{{12}}\right)}}}}$

plug in the values evaluated above,

$\displaystyle=\frac{{\frac{{-\sqrt{{{2}}}{\left(\sqrt{{{3}}}+{1}\right)}}}{{4}}}}{{\frac{{\sqrt{{{2}}}-\sqrt{{{6}}}}}{{4}}}}$

$\displaystyle=\frac{{-\sqrt{{{2}}}{\left(\sqrt{{{3}}}+{1}\right)}}}{{\sqrt{{{2}}}-\sqrt{{{6}}}}}$

rationalize the denominator,

$\displaystyle=-\frac{{{\left(\sqrt{{{6}}}+\sqrt{{{2}}}\right)}{\left(\sqrt{{{2}}}+\sqrt{{{6}}}\right)}}}{{{\left(\sqrt{{{2}}}-\sqrt{{{6}}}\right)}{\left(\sqrt{{{2}}}+\sqrt{{{6}}}\right)}}}$