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

Mutantoe
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Math Solver
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    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

3 Answers3

6

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

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

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.

Chai T. Rex
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MrYouMath
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HINT:

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

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

Lorenzo Castagno
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