Let's solve the problem step by step.
We need to calculate the values of [tex]\(\arcsin \left(\frac{1}{2}\right)\)[/tex] and [tex]\(\arccos \left(\frac{1}{2}\right)\)[/tex], square them separately, and then sum those squares. Finally, we'll determine if their sum equals 1.
1. Calculate [tex]\(\arcsin\left(\frac{1}{2}\right)\)[/tex]:
The value of [tex]\(\arcsin \left(\frac{1}{2}\right)\)[/tex] is a specific angle whose sine is [tex]\(\frac{1}{2}\)[/tex]. This angle is [tex]\(\frac{\pi}{6}\)[/tex] radians or approximately [tex]\(0.5235987755982989\)[/tex] radians.
2. Calculate [tex]\(\arccos\left(\frac{1}{2}\right)\)[/tex]:
The value of [tex]\(\arccos \left(\frac{1}{2}\right)\)[/tex] is a specific angle whose cosine is [tex]\(\frac{1}{2}\)[/tex]. This angle is [tex]\(\frac{\pi}{3}\)[/tex] radians or approximately [tex]\(1.0471975511965979\)[/tex] radians.
3. Square these values:
- Squaring [tex]\(\arcsin \left(\frac{1}{2}\right)\)[/tex]:
[tex]\[
\arcsin^2 \left(\frac{1}{2}\right) = (0.5235987755982989)^2 \approx 0.2741556778080378
\][/tex]
- Squaring [tex]\(\arccos \left(\frac{1}{2}\right)\)[/tex]:
[tex]\[
\arccos^2 \left(\frac{1}{2}\right) = (1.0471975511965979)^2 \approx 1.0966227112321512
\][/tex]
4. Sum the squares:
[tex]\[
\arcsin^2 \left(\frac{1}{2}\right) + \arccos^2 \left(\frac{1}{2}\right) = 0.2741556778080378 + 1.0966227112321512 \approx 1.370778389040189
\][/tex]
Therefore, combining the values step-by-step, the final result we obtained is approximately [tex]\(1.370778389040189\)[/tex]. This shows that the squared terms, when summed together, do not precisely equal 1 but rather approximately [tex]\(1.370778389040189\)[/tex].
