Blog: Angle x is acute and is such that tan x = sqrt(2)/4 (a) Show clearly that the exact value of sin x is 1/3 (b) Hence show that sin 2x = (4/9)sqrt(2)

Friday, October 16, 2009

Angle x is acute such that

$\displaystyle{\tan{{\left({x}\right)}}}=\frac{\sqrt{{{2}}}}{{4}}.$

Thus we have a right triangle with legs of length sqrt(2) and 4. The hypotenuse of the triangle can be found using the Pythagorean theorem:

$\displaystyle{{h}}^{{2}}={{\left(\sqrt{{{2}}}\right)}}^{{2}}+{{4}}^{{2}}$

$\displaystyle{{h}}^{{2}}={18}$

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

To find the sine, we take the ratio of the side opposite x to the hypotenuse:

$\displaystyle{\sin{{\left({x}\right)}}}=\frac{\sqrt{{{2}}}}{{3}}\sqrt{{{2}}}$

$\displaystyle=\frac{{1}}{{3}}$

Also, cos(x) is the ratio of the side adjacent to the hypotenuse so:

$\displaystyle{\cos{{\left({x}\right)}}}=\frac{{4}}{{{3}\sqrt{{{2}}}}}={2}\frac{\sqrt{{{2}}}}{{3}}$

The $\displaystyle{\sin{{\left({2}{A}\right)}}}={2}{\sin{{\left({A}\right)}}}{\cos{{\left({A}\right)}}}$ so:

$\displaystyle{\sin{{\left({2}{x}\right)}}}={2}{\left(\frac{{1}}{{3}}\right)}{\left(\frac{{{2}\sqrt{{{2}}}}}{{3}}\right)}$

$\displaystyle=\frac{{2}}{{3}}\cdot\frac{{2}}{{3}}\cdot\sqrt{{{2}}}$

$\displaystyle=\frac{{4}}{{9}}\cdot\sqrt{{{2}}}$

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