`-pi/12 = pi/6 - pi/4` Find the exact values of the sine, cosine, and tangent of the angle.

Monday, December 6, 2010

$\displaystyle-\frac{\pi}{{12}}=\frac{\pi}{{6}}-\frac{\pi}{{4}}$ Find the exact values of the sine, cosine, and tangent of the angle.

You need to find the values of sine and cosine of $\displaystyle-\frac{\pi}{{12}}$ , using the formulas $\displaystyle{\sin{{\left({a}-{b}\right)}}}={\sin{{a}}}\cdot{\cos{{b}}}-{\sin{{b}}}\cdot{\cos{{a}}}$ and $\displaystyle{\cos{{\left({a}-{b}\right)}}}={\cos{{a}}}\cdot{\cos{{b}}}+{\sin{{a}}}\cdot{\sin{{b}}}$ , such that:

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

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

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

$\displaystyle{\cos{{\left(-\frac{\pi}{{12}}\right)}}}={\cos{{\left(\frac{\pi}{{6}}-\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{6}}\right)}}}{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{\sin{{\left(\frac{\pi}{{4}}\right)}}}{\sin{{\left(\frac{\pi}{{6}}\right)}}}$

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

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

You need to evaluate tangent function such that:

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

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

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

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

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Monday, December 6, 2010

$\displaystyle-\frac{\pi}{{12}}=\frac{\pi}{{6}}-\frac{\pi}{{4}}$ Find the exact values of the sine, cosine, and tangent of the angle.

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