Express $\arcsin(x)$ in terms of $\arccos(x)$. Solve the equation 2 arctan x=arcsin x + arccos x

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: $$\sin\left(\frac{\pi}{2}-x\right) = \cos(x)$$ $$\sin(x) = \frac{\pi}{2}-\cos(x)$$

• 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. – Math Solver Sep 19 '15 at 17:08

HINT:

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

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

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

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., $$u+v=\dfrac{\pi}{2}.$$