Evaluate arccos(sqrt(2)/2) without a calculator, giving the answer as a multiple of pi. State whether the sign is + or -.

Step-by-step solution

1. 1

   Restate the question as an equation. Finding $\cos^{-1}\!\bigl(\tfrac{\sqrt2}{2}\bigr)$ means finding the angle $\theta$ with

   $\cos\theta = \frac{\sqrt{2}}{2}, \qquad 0 \le \theta \le \pi$

   That range restriction is what makes the inverse cosine a genuine function with one output.

2. 2

   Recognise the special value. $\tfrac{\sqrt2}{2}$ is the same number as $\tfrac{1}{\sqrt2} \approx 0.7071$, and it is the cosine of $45^{\circ}$ — it comes from the isosceles right triangle with legs $1$, $1$ and hypotenuse $\sqrt2$, where the adjacent side over the hypotenuse is $1/\sqrt2$. No calculator is needed.

3. 3

   Convert $45^{\circ}$ to radians. Multiply by $\pi/180$:

   $45^{\circ} \times \frac{\pi}{180} = \frac{\pi}{4}$

4. 4

   Settle the sign. The output must lie in $[0,\pi]$, and the input $\tfrac{\sqrt2}{2}$ is positive, so the angle is in the first quadrant. The answer is therefore $+\tfrac{\pi}{4}$. Note $-\tfrac{\pi}{4}$ also satisfies $\cos\theta = \tfrac{\sqrt2}{2}$, but it falls outside the principal range, so inverse cosine never returns it.

5. 5

   Verify. $\cos\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}$ ✓, and $\tfrac{\pi}{4} \approx 0.7854$ lies inside $[0,\pi]$ ✓. Had the input been $-\tfrac{\sqrt2}{2}$, the answer would instead have been $\tfrac{3\pi}{4}$, still positive — inverse cosine is never negative.