`285^@` Find the exact values of the sine, cosine, and tangent of the angle.

Sunday, January 30, 2011

$\displaystyle{{285}}^{{0}}={{270}}^{{0}}+{{15}}^{{0}}$

$\displaystyle{\sin{{\left({{270}}^{{0}}+{{15}}^{{0}}\right)}}}=-{{\cos{{15}}}}^{{0}}$

$\displaystyle=-{\cos{{\left({{45}}^{{0}}-{{30}}^{{0}}\right)}}}$

$\displaystyle=-{\left[{{\cos{{45}}}}^{{0}}\cdot{{\cos{{30}}}}^{{0}}+{{\sin{{45}}}}^{{0}}\cdot{{\sin{{30}}}}^{{0}}\right]}$

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

$\displaystyle=-{\left[\frac{{\sqrt{{3}}+{1}}}{{{2}\sqrt{{2}}}}\right]}$

$\displaystyle{\cos{{\left({{270}}^{{0}}+{{15}}^{{0}}\right)}}}={{\sin{{15}}}}^{{0}}$

$\displaystyle={\sin{{\left({{45}}^{{0}}-{{30}}^{{0}}\right)}}}$

$\displaystyle={{\sin{{45}}}}^{{0}}\cdot{{\cos{{30}}}}^{{0}}-{{\cos{{45}}}}^{{0}}\cdot{{\sin{{30}}}}^{{0}}$

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

Now $\displaystyle{\tan{{\left({{285}}^{{0}}\right)}}}$ . this can be evaluated by using $\displaystyle{\tan{{\left({A}+{B}\right)}}}$ identity or

$\displaystyle{\tan{{A}}}=\frac{{{\sin{{A}}}}}{{{\cos{{A}}}}}$

$\displaystyle{\tan{{\left({{285}}^{{0}}\right)}}}=\frac{{{\sin{{\left({{285}}^{{0}}\right)}}}}}{{{\cos{{\left({{285}}^{{0}}\right)}}}}}$

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

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

by rationalizing the denominator we get

$\displaystyle{\tan{{\left({{285}}^{{0}}\right)}}}=-{\left({2}+\sqrt{{3}}\right)}$

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