Question

 If \[\cos x = \frac{4}{5}\]  and x is acute, find tan 2x 

 

Answer 1

\[\cos x = \frac{4}{5}\]

\[\therefore \text{ sin } x = \sqrt{1 - \cos^2 x}\]
\[ = \sqrt{1 - \left( \frac{4}{5} \right)^2}\]
\[ = \sqrt{1 - \frac{16}{25}}\]
\[ = \sqrt{\frac{25 - 16}{25}}\]
\[ = \sqrt{\frac{9}{25}}\]
\[ = \frac{3}{5}\]

\[\therefore \tan x = \frac{\sin x}{\cos x}\]
\[ = \frac{\frac{3}{5}}{\frac{4}{5}}\]
\[ = \frac{3}{4}\]

Now,

\[\tan 2x = \frac{2 \text{ tan } x}{1 - \tan^2 x}\]
\[ = \frac{2\left( \frac{3}{4} \right)}{1 - \left( \frac{3}{4} \right)^2}\]
\[ = \frac{2\left( \frac{3}{4} \right)}{1 - \frac{9}{16}}\]
\[ = \frac{\frac{3}{2}}{\frac{7}{16}}\]
\[ = \frac{24}{7}\]

Hence, the value of tan 2x is \[\frac{24}{7}\] .