TrigonometrySolutions for arcsin(x) + arcsin(k) = π/2

sweatingbear

Member
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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MarkFL

Staff member
I would consider the identity:

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

Hence:

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

From this you will get the desired result...

sweatingbear

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

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

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 $$\displaystyle \frac {\pi}2$$ we must require that

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

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

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

Opalg

MHB Oldtimer
Staff member
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 $$\displaystyle 0 \leq \arcsin (k) \leq \pi$$. But since $$\displaystyle \arcsin (k)$$ cannot be any greater than $$\displaystyle \frac{\pi}2$$, we must require that $$\displaystyle 0 \leq \arcsin (k) \leq \frac{\pi}2$$. Consequently, since $$\displaystyle \arcsin (k)$$ is injective in that particular interval, we can finally arrive at $$\displaystyle 0 \leq k \leq 1$$ (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 $$\displaystyle \arcsin (k)$$ must be non-negative, which means that $0\leqslant k\leqslant1$.

sweatingbear

Member
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 $$\displaystyle \frac {\pi}2$$.

sweatingbear

Member
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

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

where

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

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

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

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

QED.

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