Why $\arccos(\frac{1}{\sqrt{5}})+\arcsin(\frac{3}{\sqrt{10}}) \neq \frac{\pi}{4}$

Question

So I want to figure out the value of $\arccos(\frac{1}{\sqrt{5}})+\arcsin(\frac{3}{\sqrt{10}})$.

Let $\alpha=\arccos(\frac{1}{\sqrt{5}})$ and $\beta=\arcsin(\frac{3}{\sqrt{10}})$. So $\alpha+\beta=\arccos(\frac{1}{\sqrt{5}})+\arcsin(\frac{3}{\sqrt{10}})$. I have tried using the sum rule for sine:

$$\sin(\alpha+\beta)=\sin(\alpha) \cos(\beta)+\cos(\alpha) \sin(\beta)$$ Furthermore we have:

$$\sin(\alpha)=\sin(\arccos(\frac{1}{\sqrt{5}}))=\frac{2}{\sqrt{5}} \\ \\sin(\beta)=\sin(\arcsin(\frac{3}{\sqrt{10}}))=\frac{3}{\sqrt{10}} \\ \cos(\alpha)=\cos(\arccos(\frac{1}{\sqrt{5}}))=\frac{1}{\sqrt{5}} \\ \cos(\beta)=\cos(\arcsin(\frac{3}{\sqrt{10}}))=\frac{1}{\sqrt{10}}$$

So $$\sin(\alpha+\beta)=\sin(\alpha) \cos(\beta)+\cos(\alpha) \sin(\beta)=\frac{1}{\sqrt{2}} \\ \implies \\ \alpha+\beta= \frac{\pi}{4} + k*2\pi \ \lor \alpha+\beta= \frac{3\pi}{4} + k*2\pi, k \in \mathbb Z$$

Since both arcsin and arccos takes positive values we get:

$$0<\alpha<\frac{\pi}{2} \ \& \ 0< \beta<\frac{\pi}{2} \implies 0<0<\alpha+\beta<\pi $$

The case $\alpha+\beta= \frac{\pi}{4} + k*2\pi$ gives: $$0<\frac{\pi}{4} + k*2\pi<pi \\ 0< \frac{1}{8}+k<\frac{1}{2} \\ -\frac{1}{8}<k<\frac{3}{8} \implies k = 0; \text{Thus we have that } \alpha+\beta = \frac{\pi}{4}$$

The case $\alpha+\beta= \frac{3\pi}{4} + k*2\pi$ gives with the same type of reasoning that $\alpha+\beta=\frac{3\pi}{4}$.

So we get the answers $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, but $\frac{\pi}{4}$ is false and I cannot not see why. What am I doing wrong?

Answer 1

Similarly,

$$\begin{align*} \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\ &= \frac{1}{\sqrt5} \frac{1}{\sqrt{10}}-\frac{2}{\sqrt5} \frac{3}{\sqrt{10}}\\ &= -\frac{1}{\sqrt2} \end{align*}$$

With the inequality that $0<\alpha + \beta < \pi$, these criteria limit $\alpha+\beta$ to exactly one possibility $\frac{3\pi}{4}$.

I guess the lesson here is to pick the more helpful trigonometric value to limit possibilities. Here, $\cos(\alpha+\beta)$ is injective for the interval $0<\alpha + \beta < \pi$.

Answer 2

The answer is $\frac\pi4$ or $\frac{3\pi}4$, not $\frac\pi4$ and $\frac{3\pi}4$, simply.

You introduce an extra solution when you take the sine of the equation.

Answer 3

Your math is correct up until the point you get

$$\sin(\alpha+\beta)=\sin(\alpha) \cos(\beta)+\cos(\alpha) \sin(\beta)=\frac{1}{\sqrt{2}}$$

What you're missing is that there are two angles on the unit circle whose sine is $\frac{1}{\sqrt{2}}$: $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ (plus any integer multiple of $2\pi$). To determine which one to use, you need to consider the range of $\alpha$ and $\beta$. Comparing with the nearest “special” angles:

• $\alpha = \arccos(\frac{1}{\sqrt{5}} \approx 0.447)$ is between $\arccos(\frac{1}{2}) = \frac{\pi}{3}$ and $\arccos(0) = \frac{\pi}{2}$.
• $\beta = \arcsin(\frac{3}{\sqrt{10}} \approx 0.948)$ is between $\arcsin(\frac{\sqrt{3}}{2} \approx 0.866) = \frac{\pi}{3}$ and $\arcsin(1) = \frac{\pi}{2}$.

Thus, $\frac{2\pi}{3} < \alpha + \beta < \pi$, so $\frac{\pi}{4} < \frac{2\pi}{3}$ is not a possible value for the sum of the angles, but $\frac{3\pi}{4}$ is.

Answer 4

You were completely OK up to and including this statement:

$$ \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) = \frac1{\sqrt2}. $$

Your conclusion is kinda sorta correct, but ambiguous:

$$ \alpha+\beta = \frac\pi4 + k2\pi\ \lor\ \alpha+\beta = \frac{3\pi}{4} + k2\pi, k \in \mathbb Z. $$

What do you mean by $k \in \mathbb Z$? Do you mean that the entire line before the comma is true for every integer $k$? Or do you mean that there is at least one integer $k$ that makes the part before the comma true? Or perhaps you meant that either the second half of the $\lor$ statement is satisfied for some integer $k$ (or all integers $k$) and that the first half is true without the condition that $k$ is an integer.

I think you meant that the entire $\lor$ statement is true for some integer $k$, that is,

$$ (\exists k \in \mathbb Z) \left(\alpha+\beta = \frac\pi4 + k2\pi \ \lor\ \alpha+\beta = \frac{3\pi}{4} + k2\pi\right). $$

The next thing you do with this, however, is to split it into two cases, so you're really using the equivalent fact $$ (\exists k \in \mathbb Z)\left(\alpha+\beta = \frac\pi4 + k2\pi\right) \ \lor\ (\exists k \in \mathbb Z)\left(\alpha+\beta = \frac{3\pi}{4} + k2\pi\right). $$

Now, if this disjunction is satisfied by the first case, that is, if $$ (\exists k \in \mathbb Z)\left(\alpha+\beta = \frac\pi4 + k2\pi\right), $$ then you have found that the value of $k$ must be $0$.

Likewise, if the disjunction is satisfied by the second case, that is, if $$ (\exists k \in \mathbb Z)\left(\alpha+\beta = \frac{3\pi}{4} + k2\pi\right), $$ then you have found that the value of $k$ must be $0$.

What you did not show is that the first case is true. You also did not show that the second case is true. But you know that at least one of these cases is true, so you have that at least one of the following is a true statement:

$$ \alpha+\beta = \frac\pi4 \quad\text{is a solution but}\quad \alpha+\beta = \frac{3\pi}{4} \quad\text{is not.} $$ $$ \alpha+\beta = \frac{3\pi}{4} \quad\text{is a solution but}\quad \alpha+\beta = \frac\pi4 \quad\text{is not.} $$ $$ \alpha+\beta = \frac{3\pi}{4} \quad\text{and}\quad \alpha+\beta = \frac\pi4 \quad\text{both are possible solutions.} $$

You need to do more work to figure out which of these is the true statement. You cannot just accept the third statement as the true one based on what you did.

In fact, it is easy to show that $\alpha > \frac\pi4$ and $\beta > \frac\pi4$, which eliminates two of the three statements in the list above and tells you which one is the true statement.