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

if $\mathrm{cos}45=\frac{1}{\sqrt{2}}$, find $\mathrm{cos}315,\mathrm{sin}270,\mathrm{sin}210,\mathrm{tan}210$
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$\mathrm{cos}45=\frac{1}{\sqrt{2}},\mathrm{sin}45=\sqrt{1-{\mathrm{cos}}^{2}\left(45\right)}=\sqrt{1-\frac{1}{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\mathrm{cos}45,$
$\mathrm{sin}90=\mathrm{sin}\left(2\cdot 45\right)=2\mathrm{sin}45\mathrm{cos}45=2\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=1,\mathrm{cos}90=\mathrm{cos}\left(2\cdot 45\right)=\mathrm{cos}45\mathrm{cos}45-\mathrm{sin}45\mathrm{sin}45=\frac{1}{2}-\frac{1}{2}=0,$
$\mathrm{sin}180=\mathrm{sin}\left(2\cdot 90\right)=2\mathrm{sin}90\mathrm{cos}90=0,\mathrm{cos}180=\mathrm{cos}\left(2\cdot 90\right)=\mathrm{cos}90\mathrm{cos}90-\mathrm{sin}90\mathrm{sin}90=0-1=-1,$
$\mathrm{sin}270=\mathrm{sin}\left(180+90\right)=\mathrm{sin}180\mathrm{cos}90+\mathrm{cos}180\mathrm{sin}90=0-1=-1.$
$\mathrm{sin}3X=\mathrm{sin}\left(2X+X\right)=\mathrm{sin}2X\mathrm{cos}X+\mathrm{cos}2X\mathrm{sin}X=2\mathrm{sin}X{\mathrm{cos}}^{2}\left(X\right)+{\mathrm{cos}}^{2}\left(X\right)\mathrm{sin}\left(X\right)-{\mathrm{sin}}^{3}\left(X\right),so\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}3X=90:$