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

sweatingbear

Member
May 3, 2013
91
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

Administrator
Staff member
Feb 24, 2012
13,775
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
May 3, 2013
91
@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
Feb 7, 2012
2,725
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
May 3, 2013
91
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
May 3, 2013
91
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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