Notes: `(7pi)/12 = pi/3 + pi/4` Find the exact values of the sine, cosine, and tangent of the angle.

Sunday, 26 June 2016

$\displaystyle\frac{{{7}\pi}}{{12}}=\frac{\pi}{{3}}+\frac{\pi}{{4}}$ Find the exact values of the sine, cosine, and tangent of the angle.

You need to evaluate the sine of $\displaystyle\frac{{{7}\pi}}{{12}}$ , using the formula $\displaystyle{\sin{{\left({a}+{b}\right)}}}={\sin{{a}}}\cdot{\cos{{b}}}+{\sin{{b}}}\cdot{\cos{{a}}}$ such that:

$\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}={\sin{{\left(\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}={\sin{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{\sin{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{3}}\right)}}}$

$\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{3}}}}{{2}}\cdot\frac{{\sqrt{{2}}}}{{2}}+\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{1}}{{2}}$

$\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{\sqrt{{3}}+{1}}}{{2}}$

You need to evaluate the cosine of $\displaystyle\frac{{{7}\pi}}{{12}}$ , using the formula $\displaystyle{\cos{{\left({a}+{b}\right)}}}={\cos{{a}}}\cdot{\cos{{b}}}-{\sin{{b}}}\cdot{\sin{{a}}}$ such that:

$\displaystyle{\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}={\cos{{\left(\frac{\pi}{{3}}+\ldots\right.}}}$

$\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{3}}}}{{2}}\cdot\frac{{\sqrt{{2}}}}{{2}}+\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{1}}{{2}}$

$\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{\sqrt{{3}}+{1}}}{{2}}$

$\displaystyle{\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}={\cos{{\left(\frac{\pi}{{3}}+\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{4}}\right)}}}-{\sin{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{3}}\right)}}}$

$\displaystyle{\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{1}}{{2}}\cdot\frac{{\sqrt{{2}}}}{{2}}-\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{\sqrt{{3}}}}{{2}}$

$\displaystyle{\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{{1}-\sqrt{{3}}}}{{2}}$

You need to evaluate the tangent of $\displaystyle\frac{{{7}\pi}}{{12}}$ , such that:

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

$\displaystyle{\tan{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{\sqrt{{3}}+{1}}}{{2}}}}{{\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{{1}-\sqrt{{3}}}}{{2}}}}$

$\displaystyle{\tan{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{3}}+{1}}}{{{1}-\sqrt{{3}}}}$

$\displaystyle{\tan{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{{\left(\sqrt{{3}}+{1}\right)}\cdot{\left({1}+\sqrt{{3}}\right)}}}{{{1}-{3}}}$

$\displaystyle{\tan{{\left(\frac{{{7}\pi}}{{12}}\right)}}}{\)}=-\frac{{{{\left(\sqrt{{3}}+{1}\right)}}^{{2}}}}{{2}}$

Hence, evaluating the sine, cosine and tangent of $\displaystyle\frac{{\tan{{\left({7}\pi\right)}}}}{{12}}$ , yields $\displaystyle{\sin{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{\sqrt{{3}}+{1}}}{{2}},{\cos{{\left(\frac{{{7}\pi}}{{12}}\right)}}}=\frac{{\sqrt{{2}}}}{{2}}\cdot\frac{{{1}-\sqrt{{3}}}}{{2}},\frac{{\tan{{\left({7}\pi\right)}}}}{{12}}=-\frac{{{{\left(\sqrt{{3}}+{1}\right)}}^{{2}}}}{{2}}.$