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

Thursday, 10 December 2015

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

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

As we know that $\displaystyle{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={\cos{{\left(\theta\right)}}}$

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

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

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

$\displaystyle{\sin{{\left(\frac{{{13}\pi}}{{12}}\right)}}}={\left({\left(\frac{{1}}{{2}}\right)}{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}-{\left({\left(\frac{\sqrt{{{3}}}}{{2}}\right)}{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}\right)}\right.}$

$\displaystyle{\sin{{\left(\frac{{{13}\pi}}{{12}}\right)}}}=\frac{{{1}-\sqrt{{{3}}}}}{{{2}\sqrt{{{2}}}}}$

$\displaystyle{\sin{{\left(\frac{{{13}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{{2}}}-\sqrt{{{6}}}}}{{4}}$

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

We know that $\displaystyle{\cos{{\left(\frac{\pi}{{2}}+\theta\right)}}}=-{\sin{{\left(\theta\right)}}}$

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

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

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

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

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

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

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

plug in the values evaluated above,

$\displaystyle{\tan{{\left(\frac{{{13}\pi}}{{12}}\right)}}}=\frac{{\frac{{\sqrt{{{2}}}-\sqrt{{{6}}}}}{{4}}}}{{-\frac{{\sqrt{{{6}}}+\sqrt{{{2}}}}}{{4}}}}$

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

rationalizing the denominator,

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

$\displaystyle=\frac{{\sqrt{{{12}}}-{2}-{6}+\sqrt{{{12}}}}}{{-{4}}}$

$\displaystyle=\frac{{{2}\sqrt{{{12}}}-{8}}}{{-{4}}}$

$\displaystyle=\frac{{{2}\cdot{2}\sqrt{{{3}}}-{8}}}{{-{4}}}$

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