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#### sweatingbear

##### Member

- May 3, 2013

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**Q:**For which values of \(\displaystyle k\) is the equation

\(\displaystyle \arcsin (x) + \arcsin (k) = \frac {\pi}2 \, ,\)

solvable? Furthermore, find an expression for explicit solution without arcus functions.

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**S:**Let us first find the explicit solution:

\(\displaystyle \arcsin (x) + \arcsin (k) = \frac {\pi}2 \ \Longleftrightarrow \ \arcsin (x) = \frac {\pi}2 - \arcsin (k) \, .\)

Since \(\displaystyle \arcsin (x)\) is injective in its domain, \(\displaystyle x\) must consequently equal the sine of \(\displaystyle \frac {\pi}2 - \arcsin (k)\).

\(\displaystyle x = \sin \left[ \frac {\pi}2 - \arcsin (k) \right] \, .\)

The subtraction formulae yield

\(\displaystyle x = \underbrace{\sin \left( \frac {\pi}2 \right)}_{1}\cos \left[ \arcsin (k) \right] - \sin \left[ \arcsin (k) \right]\underbrace{\cos \left( \frac {\pi}2 \right)}_{0} \ \Longleftrightarrow \ x = \cos \left[ \arcsin (k) \right] \, .\)

We can derive \(\displaystyle \cos \left[\arcsin (k)\right] = \sqrt{1 - k^2}\) by building right triangles and thus have

\(\displaystyle x = \sqrt{1 - k^2} \, .\)

In terms of finding an explicit solution for \(\displaystyle x\) without arcus functions, it is correct. However we need to determine the interval of \(\displaystyle k\). Looking at the argument of the square root one can conclude that, in order for \(\displaystyle x \in \mathbb{R}\), we must have that \(\displaystyle -1 \leqslant k \leqslant 1\). Surprisingly, it turns out that this interval is incorrect.

According to the key in my book, one must have that \(\displaystyle 0 \leqslant k \leqslant 1\) (and it makes sense, because if we plug in \(\displaystyle k = -1\) in the given equation we arrive at an absurd relation). This is where I am stuck: I do not know where in my argument I can algebraically arrive at \(\displaystyle k \geqslant 0\).

Here's a spontaneous thought: The domain of cosine is principally \(\displaystyle \left[0, \pi\right]\) and therefore if \(\displaystyle x = \cos \left[ \arcsin (k) \right]\) then \(\displaystyle 0 \leqslant \arcsin (k) \leqslant \pi\). But since \(\displaystyle \arcsin (k)\) cannot be any greater than \(\displaystyle \frac{\pi}2\), we must require that \(\displaystyle 0 \leqslant \arcsin (k) \leqslant \frac{\pi}2\). Consequently, since \(\displaystyle \arcsin (k)\) is injective in that particular interval, we can finally arrive at \(\displaystyle 0 \leqslant k \leqslant 1\) (taking the sine of every side of inequality).

But I am still insecure about this particular method and train of thought. Anyone got a better suggestion?

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