MHB Solutions for arcsin(x) + arcsin(k) = π/2

[SweatingBear]

Q: For which values of $$k$$ is the equation

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

solvable? Furthermore, find an expression for explicit solution without arcus functions.
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S: Let us first find the explicit solution:

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

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

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

The subtraction formulae yield

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

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

$$x = \sqrt{1 - k^2} \, .$$

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

According to the key in my book, one must have that $$0 \leqslant k \leqslant 1$$ (and it makes sense, because if we plug in $$k = -1$$ 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 $$k \geqslant 0$$.

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

 

[MarkFL]

I would consider the identity:

$$\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$$

Hence:

$$\sin^{-1}(k)=\cos^{-1}(x)$$

From this you will get the desired result...

 

[SweatingBear]

@MarkFL: Ok, let us try that. For $$g(x) = \arccos (x)$$ we have that the range is $$0 \leq \arccos (x) \leq \pi$$. Since we are equating it to $$\arcsin (k)$$, one must require that

$$0 \leq \arcsin (k) \leq \pi \, .$$

This leads me back to the same line of argument as in my previous post. Since the inverse sine function cannot be any greater than $$\frac {\pi}2$$ we must require that

$$0 \leq \arcsin (k) \leq \frac {\pi}2 \, ,$$

and consequently have that $$0 \leq k \leq 1$$.

Does this mean that the argument I made in my previous post about the domain of the cosine function is valid?

 

[Opalg]

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

That line of reasoning looks absolutely correct. The range of the $\arcsin$ function is the interval $[-\pi/2,\pi/2]$. If two numbers in that interval have sum $\pi/2$ then they must both be non-negative. Therefore $$\arcsin (k)$$ must be non-negative, which means that $0\leqslant k\leqslant1$.

 

[SweatingBear]

That line of reasoning looks absolutely correct.

Great, thank you for the assessment.

The range of the $\arcsin$ function is the interval $[-\pi/2,\pi/2]$. If two numbers in that interval have sum $\pi/2$ then they must both be non-negative.

My train of thoughts was on that track as well, however I did not find it to be sufficiently rigorous. As a matter of fact, I am struggling with wrapping my mind around that: Why do both terms necessarily have to be non-negative? Would it not be possible for the other number to be negative, the other one to be positive and their sum to still equal $$\frac {\pi}2$$.

 

[SweatingBear]

On second thoughts, you are right; the both terms definitely need to be non-negative. Here is an idea for a proof.

Proof:

Suppose we have

$$p + q = \frac {\pi}2 \, ,$$

where

$$-\frac{\pi}2 \leqslant p, \, q \leqslant \frac{\pi}2 \ \ \wedge \ \ p, \, q \in \mathbb{R} \, . $$

Suppose either $$q$$ or $$p$$ is negative (does not matter which one, I choose $$q$$); then we can let $$q = -k$$ where $$k \in \mathbb{R}$$. Thus

$$p + (-k) = \frac {\pi}2 \ \Longleftrightarrow \ p = \frac {\pi}2 + k \, .$$

If it is the case that $$k > 0$$ then we have a contradiction because then $$p$$ would equal $$\frac {\pi}2$$ plus a small positive addition of $$k$$, which contradicts the requirement of $$p$$ maximally equalling $$\frac {\pi}2$$. The case $$k = 0$$ does not lead to a contradiction since $$p$$ is allowed to equal $$\frac {\pi}2$$. Thus $$p$$ and $$q$$ must be non-negative.

QED.