# if cos 45=1/sqrt2, find cos 315, sin 270, sin 210, tan 210

Question
if $$\displaystyle{\cos{{45}}}=\frac{{1}}{\sqrt{{2}}}$$, find $$\displaystyle{\cos{{315}}},{\sin{{270}}},{\sin{{210}}},{\tan{{210}}}$$

2021-01-26
$$\displaystyle{\cos{{45}}}=\frac{{1}}{\sqrt{{2}}},{\sin{{45}}}=\sqrt{{{1}-{{\cos}^{{2}}{\left({45}\right)}}}}=\sqrt{{{1}-\frac{{1}}{{2}}}}=\sqrt{{\frac{{1}}{{2}}}}=\frac{{1}}{\sqrt{{2}}}={\cos{{45}}},$$
$$\displaystyle{\sin{{90}}}={\sin{{\left({2}\cdot{45}\right)}}}={2}{\sin{{45}}}{\cos{{45}}}={2}\cdot\frac{{1}}{\sqrt{{2}}}\cdot\frac{{1}}{\sqrt{{2}}}={1},{\cos{{90}}}={\cos{{\left({2}\cdot{45}\right)}}}={\cos{{45}}}{\cos{{45}}}-{\sin{{45}}}{\sin{{45}}}=\frac{{1}}{{2}}-\frac{{1}}{{2}}={0},$$
$$\displaystyle{\sin{{180}}}={\sin{{\left({2}\cdot{90}\right)}}}={2}{\sin{{90}}}{\cos{{90}}}={0},{\cos{{180}}}={\cos{{\left({2}\cdot{90}\right)}}}={\cos{{90}}}{\cos{{90}}}-{\sin{{90}}}{\sin{{90}}}={0}-{1}=-{1},$$
$$\displaystyle{\cos{{\left({X}\right)}}}={\cos{{\left(-{X}\right)}}},{s}{o}{\cos{{45}}}={\cos{{\left(-{45}\right)}}}={\cos{{\left({360}-{45}\right)}}}={\cos{{\left({315}\right)}}}={1}\sqrt{{2}}.$$
$$\displaystyle{\sin{{270}}}={\sin{{\left({180}+{90}\right)}}}={\sin{{180}}}{\cos{{90}}}+{\cos{{180}}}{\sin{{90}}}={0}-{1}=-{1}.$$
$$\displaystyle{\sin{{3}}}{X}={\sin{{\left({2}{X}+{X}\right)}}}={\sin{{2}}}{X}{\cos{{X}}}+{\cos{{2}}}{X}{\sin{{X}}}={2}{\sin{{X}}}{{\cos}^{{2}}{\left({X}\right)}}+{{\cos}^{{2}}{\left({X}\right)}}{\sin{{\left({X}\right)}}}-{{\sin}^{{3}}{\left({X}\right)}},{s}{o}{\quad\text{if}\quad}{3}{X}={90}:$$
$$\displaystyle{1}={2}{\sin{{30}}}{\left({1}-{{\sin}^{{2}}{\left({30}\right)}}\right)}+{\left({1}-{{\sin}^{{2}}{\left({30}\right)}}\right)}{\sin{{\left({30}\right)}}}-{{\sin}^{{3}}{\left({30}\right)}}={2}{\sin{{30}}}-{2}{{\sin}^{{3}}{\left({30}\right)}}+{\sin{{\left({30}\right)}}}-{{\sin}^{{3}}{\left({30}\right)}}-{{\sin}^{{3}}{\left({30}\right)}}=$$
$$\displaystyle{3}{\sin{{30}}}-{4}{{\sin}^{{{30}}},}{s}{o}{4}{{\sin}^{{3}}{\left({30}\right)}}-{3}{\sin{{30}}}+{1}={0}={\left({2}{\sin{{30}}}-{1}\right)}^{{2}}{\left({\sin{{30}}}+{1}\right)},{s}{o}{\sin{{30}}}=\frac{{1}}{{2}}{\quad\text{or}\quad}-{1},{b}{u}{t}{\sin{{270}}}=-{1}{s}{o}{\sin{{30}}}=\frac{{1}}{{2}}.$$
$$\displaystyle{\cos{{30}}}=\sqrt{{{1}-{{\sin}^{{2}}{\left({30}\right)}}}}=\sqrt{{\frac{{3}}{{4}}}}=\frac{\sqrt{{3}}}{{2}}.{\left({3}\cdot{270}={810}={2}\cdot{360}+{90},{\sin{{810}}}={\sin{{90}}}={1},\right.}$$ hence the -1 solution of the cubic.)
$$\displaystyle{\sin{{210}}}={\sin{{\left({180}+{30}\right)}}}={\sin{{180}}}{\cos{{30}}}+{\cos{{180}}}{\sin{{30}}}={0}-\frac{{1}}{{2}}=-\frac{{1}}{{2}}.$$
$$\displaystyle{\cos{{210}}}={\cos{{\left({180}+{30}\right)}}}={\cos{{180}}}{\cos{{30}}}-{\sin{{180}}}{\sin{{30}}}=-\frac{\sqrt{{3}}}{{2}},{\tan{{210}}}=\frac{{\sin{{210}}}}{{\cos{{210}}}}=\frac{{-\frac{{1}}{{2}}}}{{-{\sqrt{{3}}}}}/{2}{)}=\frac{{1}}{\sqrt{{3}}}.$$

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