Express $\arcsin(x)$ in terms of $\arccos(x)$.

Using the same, solve the equation

$$ 2\,\tan^{-1}x = \sin^{-1} x + \cos^{-1} x $$

I'm not sure if I am on the right track, but here is what i did: $$\sin\left(\frac{\pi}{2}-x\right) = \cos(x)$$ $$\sin(x) = \frac{\pi}{2}-\cos(x)$$

  • $\begingroup$ The answer for first one is arcsin x =pi/2-arccos x, or sin^-1(x)=pi/2-cos^-1(x). And for the equation, it is 1. Someone please guide me through. $\endgroup$ – Math Solver Sep 19 '15 at 17:08


You don't need too much hassle as by definition, $$\sin^{-1} x + \cos^{-1} x=\dfrac\pi2 $$ for $-1\le x\le1$

  • $\begingroup$ Yes, i got part one. But now how will I solve the equation given above? $\endgroup$ – Math Solver Sep 20 '15 at 4:09
  • $\begingroup$ @MathSolver, So, we have $\arctan x=\dfrac\pi4\implies x=\tan\dfrac\pi4$ $\endgroup$ – lab bhattacharjee Sep 20 '15 at 15:51

$$x = \sin(y)$$ $$x = \cos(\frac{\pi}{2}-y)$$

$$y=\arcsin(x)$$ $$\frac{\pi}{2}-y=\arccos(x)$$

Adding the last equations will give your identity: $$\frac{\pi}{2}=\arcsin(x)+\arccos(x)$$

Now you can solve the equation: $$2\arctan(x)=\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$ $$\arctan(x)=\frac{\pi}{4}$$

I leave the rest to you.


Solve the equation:

You can solve the above assuming that (I'll provide a proof below):


Starting from this, we can add to the right side: $2 arctan(x)$


$arcsen(x)+arccos(x)=\dfrac{\pi}{2}=2 arctan(x)$

We deduce:

$arctan(x) =\dfrac{\pi}{4}$


$tan(arctan(x))=tan(\dfrac{\pi}{4})$ then $x=1.$

Proof of:


Considering $u= arcsen(x)$ and $v=arccos(x)$, we have that $u \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$ and $v \in [0, \pi]$ moreover:

$x = sen(u)$ and $x= cos(v)$

so that:

$sen(u)=cos(v)$ (1)

Given that: $u \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$, $v \in [0, \pi]$ and $cos(v)=sen(-v,\dfrac{\pi}{2})$ with $-v+ \dfrac{\pi}{2} u \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$, we make sure that (1) we have $u$ and $v$ such that $u=\dfrac{\pi}{2}-v$, i.e., $u+v=\dfrac{\pi}{2}.$


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.