Inverse trigonometry practice problems with handwritten solutions.

Okay, now I have a clear view of the entire worksheet. It contains 45 problems, numbered 1 through 45, arranged in two columns. The problems involve finding the exact value of expressions with inverse trigonometric functions like `arcsin`, `arccos`, and `arctan`. Many of them are compositions, such as `sin(arcsin(x))` or `cos(arccos(x))`, which often simplify directly. Others involve evaluating the inverse function at a specific value, like `arcsin(1/2)`. There's also one problem, number 45, that asks to "complete the table," but the table itself is not visible in this image. I will now begin solving these problems one by one, starting with the first one.

I will start by solving problem 1: `sin(arcsin(1/2))`.

The function `arcsin(x)` (or `sin⁻¹(x)`) is the inverse of the sine function. By definition, if `y = arcsin(x)`, then `sin(y) = x`. Therefore, for any `x` within the domain of `arcsin` (which is `[-1, 1]`), the composition `sin(arcsin(x))` will simply equal `x`.

In this case, `x = 1/2`, which is within the valid domain. So, `sin(arcsin(1/2)) = 1/2`.

Now, let's solve problem 2: `cos(arccos(√3/2))`.

Similarly, `arccos(x)` (or `cos⁻¹(x)`) is the inverse of the cosine function. If `y = arccos(x)`, then `cos(y) = x`. Thus, the composition `cos(arccos(x))` equals `x` for any `x` in the domain of `arccos` (which is also `[-1, 1]`).

Here, `x = √3/2`, which is approximately 0.866 and is within the valid domain. So, `cos(arccos(√3/2)) = √3/2`.

Next, problem 3: `tan(arctan(-1))`.

The same logic applies. `arctan(x)` (or `tan⁻¹(x)`) is the inverse of the tangent function. If `y = arctan(x)`, then `tan(y) = x`. So, `tan(arctan(x)) = x` for all real numbers `x`.

Therefore, `tan(arctan(-1)) = -1`.

Let's move on to problem 4: `sin(arcsin(-√2/2))`.

Again, using the identity `sin(arcsin(x)) = x`, we get `sin(arcsin(-√2/2)) = -√2/2`.

Problem 5: `cos(arccos(-1/2))`.

Using `cos(arccos(x)) = x`, we find `cos(arccos(-1/2)) = -1/2`.

Problem 6: `tan(arctan(√3))`.

Using `tan(arctan(x)) = x`, we get `tan(arctan(√3)) = √3`.

Problem 7: `sin(arcsin(-√3/2))`.

This follows the same pattern: `sin(arcsin(-√3/2)) = -√3/2`.

Problem 8: `cos(arccos(-√2/2))`.

And again: `cos(arccos(-√2/2)) = -√2/2`.

Problem 9: `tan(arctan(-√3/3))`.

Finally for this set: `tan(arctan(-√3/3)) = -√3/3`.

Now, let's tackle problem 10: `arcsin(1/2)`.

This is asking for the angle whose sine is `1/2`. We need to recall the unit circle. The sine of an angle is the y-coordinate on the unit circle. The angle in the range `[-π/2, π/2]` (the principal range for `arcsin`) that has a sine of `1/2` is `π/3` radians, or 60 degrees. However, since `arcsin` returns values in `[-π/2, π/2]`, and `sin(π/3) = 1/2`, the answer is `π/3`.

Wait, let me correct that. `sin(π/3) = √3/2`, not `1/2`. My mistake. The angle whose sine is `1/2` in the principal range is `π/6` (or 30 degrees). So, `arcsin(1/2) = π/6`.

Problem 11: `arccos(√3/2)`.

This asks for the angle whose cosine is `√3/2`. The cosine is the x-coordinate. In the range `[0, π]` (the principal range for `arccos`), the angle with a cosine of `√3/2` is `π/6`.

Problem 12: `arctan(1)`.

This asks for the angle whose tangent is `1`. Tangent is `sin/cos`. In the range `(-π/2, π/2)` (the principal range for `arctan`), the angle where `sin = cos` is `π/4`. So, `arctan(1) = π/4`.

Problem 13: `arcsin(-1/2)`.

This is the angle whose sine is `-1/2`. In the range `[-π/2, π/2]`, this is `-π/6`.

Problem 14: `arccos(-√2/2)`.

This is the angle whose cosine is `-√2/2`. In the range `[0, π]`, this is `3π/4`.

Problem 15: `arctan(-√3)`.

This is the angle whose tangent is `-√3`. In the range `(-π/2, π/2)`, this is `-π/3`.

Problem 16: `arcsin(-√3/2)`.

This is the angle whose sine is `-√3/2`. In the range `[-π/2, π/2]`, this is `-π/3`.

Problem 17: `arccos(-1/2)`.

This is the angle whose cosine is `-1/2`. In the range `[0, π]`, this is `2π/3`.

Problem 18: `arctan(-1)`.

This is the angle whose tangent is `-1`. In the range `(-π/2, π/2)`, this is `-π/4`.

Problem 19: `arcsin(0)`.

This is the angle whose sine is `0`. In the range `[-π/2, π/2]`, this is `0`.

Problem 20: `arccos(0)`.

This is the angle whose cosine is `0`. In the range `[0, π]`, this is `π/2`.

Problem 21: `arctan(0)`.

This is the angle whose tangent is `0`. In the range `(-π/2, π/2)`, this is `0`.

Problem 22: `arcsin(-1)`.

This is the angle whose sine is `-1`. In the range `[-π/2, π/2]`, this is `-π/2`.

Problem 23: `arccos(1)`.

This is the angle whose cosine is `1`. In the range `[0, π]`, this is `0`.

Problem 24: `arctan(∞)`.

This is a bit tricky. The tangent function approaches infinity as the angle approaches `π/2` from the left. So, `arctan(∞)` is defined as `π/2`.

Problem 25: `arcsin(1)`.

This is the angle whose sine is `1`. In the range `[-π/2, π/2]`, this is `π/2`.

Problem 26: `arccos(-1)`.

This is the angle whose cosine is `-1`. In the range `[0, π]`, this is `π`.

Problem 27: `arctan(-∞)`.

Similarly, the tangent function approaches negative infinity as the angle approaches `-π/2` from the right. So, `arctan(-∞)` is defined as `-π/2`.

Problem 28: `arcsin(√2/2)`.

This is the angle whose sine is `√2/2`. In the range `[-π/2, π/2]`, this is `π/4`.

Problem 29: `arccos(√2/2)`.

This is the angle whose cosine is `√2/2`. In the range `[0, π]`, this is `π/4`.

Problem 30: `arctan(√3)`.

This is the angle whose tangent is `√3`. In the range `(-π/2, π/2)`, this is `π/3`.

Problem 31: `arcsin(-√2/2)`.

This is the angle whose sine is `-√2/2`. In the range `[-π/2, π/2]`, this is `-π/4`.

Problem 32: `arccos(-√3/2)`.

This is the angle whose cosine is `-√3/2`. In the range `[0, π]`, this is `5π/6`.

Problem 33: `arctan(-√3/3)`.

This is the angle whose tangent is `-√3/3`. In the range `(-π/2, π/2)`, this is `-π/6`.

Problem 34: `arcsin(√3/2)`.

This is the angle whose sine is `√3/2`. In the range `[-π/2, π/2]`, this is `π/3`.

Problem 35: `arccos(-1/2)`.

We already did this: `2π/3`.

Problem 36: `arctan(√3/3)`.

This is the angle whose tangent is `√3/3`. In the range `(-π/2, π/2)`, this is `π/6`.

Problem 37: `arcsin(1/2)`.

We already did this: `π/6`.

Problem 38: `arccos(1/2)`.

This is the angle whose cosine is `1/2`. In the range `[0, π]`, this is `π/3`.

Problem 39: `arctan(-1)`.

We already did this: `-π/4`.

Problem 40: `arcsin(-1/2)`.

We already did this: `-π/6`.

Problem 41: `arccos(-√3/2)`.

We already did this: `5π/6`.

Problem 42: `arctan(1)`.

We already did this: `π/4`.

Problem 43: `arcsin(0)`.

We already did this: `0`.

Problem 44: `arccos(0)`.

We already did this: `π/2`.

Problem 45: "complete the table".

As mentioned before, the table is not visible in the image. Without seeing the table, I cannot complete it. This might be a separate part of the worksheet that was not included in the uploaded image.

To summarize, I have solved all 44 problems that are visible on the worksheet. The key to solving most of these is understanding the relationship between a trigonometric function and its inverse, and knowing the standard angles on the unit circle. For problems that are compositions like `sin(arcsin(x))`, the answer is simply `x`. For problems that ask for the inverse function of a value, you need to recall the angle in the appropriate range that produces that value.

The final answer for each problem is as follows:

1. 1/2
2. √3/2
3. -1
4. -√2/2
5. -1/2
6. √3
7. -√3/2
8. -√2/2
9. -√3/3
10. π/6
11. π/6
12. π/4
13. -π/6
14. 3π/4
15. -π/3
16. -π/3
17. 2π/3
18. -π/4
19. 0
20. π/2
21. 0
22. -π/2
23. 0
24. π/2
25. π/2
26. π
27. -π/2
28. π/4
29. π/4
30. π/3
31. -π/4
32. 5π/6
33. -π/6
34. π/3
35. 2π/3
36. π/6
37. π/6
38. π/3
39. -π/4
40. -π/6
41. 5π/6
42. π/4
43. 0
44. π/2
45. [Table not visible]

For problem 45, since the table is not shown, I am unable to provide an answer.