Q.Find the principal value of the following: $\cos^{-1} \left( -\frac{1}{\sqrt{2}} \right)$

The principal value branch of $\cos^{-1}$ is $[0, \pi]$, so we need the unique angle $\theta$ in $[0, \pi]$ with $\cos \theta = -\frac{1}{\sqrt{2}}$.

Step 1: The value is negative, and on $[0, \pi]$ cosine is negative only in the second quadrant $\left(\frac{\pi}{2}, \pi\right)$, so $\theta$ lies there.

Step 2: Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos(\pi - \alpha) = -\cos \alpha$, …

The inverse cosine function, $\cos^{-1}(x)$ or $\arccos(x)$, returns the angle whose cosine is $x$. But because cosine is not one-to-one over all real numbers, we restrict its domain to get a well-defined inverse. The standard choice for the principal value branch of $\cos^{-1}$ is the interval $[0, \pi]$.

So when we ask for the principal value of $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$, we are looking for an angle $\theta$ such that:

• $\cos \theta = -\frac{1}{\sqrt{2}}$
• $\theta \in [0, \pi]$

Now, $-\frac{1}{\sqrt{2}}$ is negative. On $[0, \pi]$, cosine is positive in $(0, \frac{\pi}{2})$ and negative in $(\frac{\pi}{2}, \pi)$. So our angle must lie in the second quadrant.

We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Using the identity $\cos(\pi - \alpha) = -\cos \alpha$, we get:

Thus $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ is a candidate. Check: $\frac{3\pi}{4}$ is indeed in $[0, \pi]$, and its cosine is $-\frac{1}{\sqrt{2}}$. …

Method: Inverse cosine of a negative number

When $\cos^{-1}$ is applied to a negative value, the answer is a second-quadrant angle; here is the reliable route.

Steps

Step 1: Fix the range

$\cos^{-1}$ returns an angle in $[0, \pi]$. A negative argument forces the answer into $\left(\frac{\pi}{2}, \pi\right]$.

Step 2: Find the reference angle

Drop the sign and recall the acute angle with that cosine magnitude (e.g. $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$). …

Common Mistakes

Mistake 1: Answering $\frac{\pi}{4}$ and ignoring the sign

Why it's wrong: $\cos\frac{\pi}{4} = +\frac{1}{\sqrt{2}}$, not $-\frac{1}{\sqrt{2}}$. Correct approach: for a negative argument use the supplement, $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Mistake 2: Writing $-\frac{\pi}{4}$ as if $\cos^{-1}$ were odd …