Evaluate arccos(0.75), giving the answer in both degrees and radians.

Step-by-step solution

1. 1

   Understand what is being asked. $\cos^{-1}(0.75)$ is the angle $\theta$ with $\cos\theta=0.75$ and $0\le\theta\le\pi$ — the principal range for inverse cosine, which unlike arcsin runs from $0$ to $180^{\circ}$ rather than $-90^{\circ}$ to $90^{\circ}$.

2. 2

   Check the input and predict the answer roughly. The domain of $\cos^{-1}$ is $[-1,1]$ and $0.75$ is inside it. Since $\cos 0^{\circ}=1$ and $\cos 60^{\circ}=0.5$, and $0.75$ sits between them, the answer must lie between $0^{\circ}$ and $60^{\circ}$ — closer to $45^{\circ}$ because $\cos 45^{\circ}\approx 0.707$. This estimate is what catches a calculator in the wrong mode.

3. 3

   Recognise that $0.75$ is not a special angle. Unlike $\tfrac12$, $\tfrac{\sqrt2}{2}$ or $\tfrac{\sqrt3}{2}$, the value $0.75$ corresponds to no standard triangle, so there is no exact closed form and a calculator is genuinely required.

4. 4

   Evaluate in degrees.

   $\cos^{-1}(0.75)=41.4096^{\circ}\approx 41.41^{\circ}$

   This sits between $0^{\circ}$ and $60^{\circ}$ and just below $45^{\circ}$, exactly as predicted.

5. 5

   Convert to radians. Multiply by $\dfrac{\pi}{180}$:

   $41.4096\times\frac{\pi}{180}=0.7227\ \text{rad}$

   As a check, $0.7227$ is a little under $\tfrac{\pi}{4}=0.7854$, consistent with the angle being just under $45^{\circ}$. Verifying back: $\cos(0.7227)=0.75000$ ✓.