0
$\begingroup$

Need help solving the problem $\tan(\tan^{-1}(\frac{2}{5}))$ using inverse functions without the use of a calculator, I have no idea how to use $\frac{2}{5}$ as I am only familiar with numbers on the unit circle.

$\endgroup$
  • 1
    $\begingroup$ What makes you think it's not $\frac25$? $\endgroup$ – vrugtehagel May 6 '17 at 23:57
  • $\begingroup$ What do you know about tan and it's inverse? $\endgroup$ – DaveNine May 7 '17 at 0:00
1
$\begingroup$

Think about what the inverse tangent function is.

In English:

$\tan^{-1} x$ takes an input number $x$ and returns a number (or angle), such that the tangent of that number is $x$. So $\tan^{-1} \left( \frac{2}{5} \right)$ will return a number that the $\tan()$ of that number is $\frac{2}{5}$. However, you're putting that number right back into the $\tan$ function ($\tan\left( \tan^{-1} x \right)$). So you're taking the tangent of a number that when input into the tangent function is $\frac{2}{5}$.

In math:

$$y = \tan x \implies x = \tan^{-1} y$$ $$ \tan \left( \tan^{-1} x \right) = x $$ In general, $f\left( f^{-1} (x) \right) = x$

$\endgroup$
1
$\begingroup$

Basically tan$^{-1}(2/5))$ is some angle who's tan value is $2/5$. So, tan(tan$^{-1}(2/5))$ is just $2/5$.

$\endgroup$
0
$\begingroup$

Hint: $$\tan^{-1}(\tan \theta)=\theta $$ for $\theta \in (-\pi/2 , \pi/2)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.