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: $$\sin\left(\frac{\pi}{2}-x\right) = \cos(x)$$ $$\sin(x) = \frac{\pi}{2}-\cos(x)$$
$\endgroup$ 13 Answers
$\begingroup$$$x = \sin(y)$$$$x = \cos\left(\frac{\pi}{2}-y\right)$$
$$y=\arcsin(x)$$$$\frac{\pi}{2}-y=\arccos(x)$$
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.
$\endgroup$ $\begingroup$HINT:
You don't need too much hassle as by definition, $$\sin^{-1} x + \cos^{-1} x=\dfrac\pi2 $$ for $-1\le x\le1$
$\endgroup$ 2 $\begingroup$Solve the equation:
You can solve the above assuming that (I'll provide a proof below):
$arcsen(x)+arccos(x)=\dfrac{\pi}{2}$
Starting from this, we can add to the right side: $2 arctan(x)$
Obtaining:
$arcsen(x)+arccos(x)=\dfrac{\pi}{2}=2 arctan(x)$
We deduce:
$arctan(x) =\dfrac{\pi}{4}$
Solving,
$tan(arctan(x))=tan(\dfrac{\pi}{4})$ then $x=1.$
Proof of:
$arcsen(x)+arccos(x)=\dfrac{\pi}{2}$
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., $u+v=\dfrac{\pi}{2}.$
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