Pre calculus worksheet with inverse trig function exercises for evaluation.

Let's solve each of these inverse trigonometric function problems without a calculator, using known values and identities.

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1. $\sin^{-1}\left(\frac{1}{2}\right)$

We are looking for the angle whose sine is $\frac{1}{2}$, within the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

$$
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \Rightarrow \sin^{-1}\left(\frac{1}{2}\right) = \boxed{\frac{\pi}{6}}
$$

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2. $\cos^{-1}\left(\frac{1}{2}\right)$

Find the angle whose cosine is $\frac{1}{2}$, in the range $[0, \pi]$.

$$
\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \Rightarrow \cos^{-1}\left(\frac{1}{2}\right) = \boxed{\frac{\pi}{3}}
$$

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3. $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

Note: $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

We know:
$$
\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \Rightarrow \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \boxed{\frac{\pi}{6}}
$$

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4. $\arccos\left(\frac{\sqrt{3}}{2}\right)$

Find angle where cosine is $\frac{\sqrt{3}}{2}$, in $[0, \pi]$.

$$
\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \Rightarrow \arccos\left(\frac{\sqrt{3}}{2}\right) = \boxed{\frac{\pi}{6}}
$$

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5. $\arctan\left(\frac{\sqrt{3}}{3}\right)$

Same as #3:
$$
\arctan\left(\frac{\sqrt{3}}{3}\right) = \boxed{\frac{\pi}{6}}
$$

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6. $\arctan(1)$

We know:
$$
\tan\left(\frac{\pi}{4}\right) = 1 \Rightarrow \arctan(1) = \boxed{\frac{\pi}{4}}
$$

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7. $\arcsin\left(-\frac{1}{2}\right)$

Sine is negative in the fourth quadrant, and $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$, and this is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

$$
\arcsin\left(-\frac{1}{2}\right) = \boxed{-\frac{\pi}{6}}
$$

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8. $\arccos\left(-\frac{1}{2}\right)$

Cosine is negative in the second quadrant. $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$, and $\frac{2\pi}{3} \in [0, \pi]$.

$$
\arccos\left(-\frac{1}{2}\right) = \boxed{\frac{2\pi}{3}}
$$

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9. $\arctan\left(-\frac{4}{3}\right)$

This one is tricky because it’s not a standard value. But we can leave it as an expression or recognize that:

The arctangent of a negative number is negative, and since $\tan^{-1}(x)$ is odd:

$$
\arctan\left(-\frac{4}{3}\right) = -\arctan\left(\frac{4}{3}\right)
$$

But since $\frac{4}{3}$ is not a standard ratio, we cannot simplify further without a calculator. However, if the problem expects an exact answer, it might be expecting us to leave it as is, or perhaps there's a typo.

Wait — let's check if this is meant to be a special angle? $\frac{4}{3}$ is not a standard tangent value.

So unless more context is given, we must say:

$$
\arctan\left(-\frac{4}{3}\right) = \boxed{-\arctan\left(\frac{4}{3}\right)}
$$

But since the instruction says "evaluate without a calculator", and no exact value exists in terms of $\pi$, maybe this is a mistake?

Alternatively, perhaps it's meant to be $\arctan\left(-\frac{\sqrt{3}}{3}\right)$, but it's written as $-\frac{4}{3}$.

Assuming it's correct as written, we cannot simplify further exactly. So we just write:

$$
\boxed{-\arctan\left(\frac{4}{3}\right)}
$$

But wait — maybe it's a trick question? Let's move on and come back.

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10. $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Cosine is negative → second quadrant.

$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, and $\frac{5\pi}{6} \in [0, \pi]$

$$
\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \boxed{\frac{5\pi}{6}}
$$

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11. $\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$

Sine is negative → fourth quadrant.

$\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$, and $-\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

$$
\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) = \boxed{-\frac{\pi}{4}}
$$

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12. $\tan^{-1}(-1)$

$\tan\left(-\frac{\pi}{4}\right) = -1$, and $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$$
\tan^{-1}(-1) = \boxed{-\frac{\pi}{4}}
$$

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13. $\sin^2(0)$

$\sin(0) = 0$, so $\sin^2(0) = 0^2 = \boxed{0}$

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14. $\cos^2(\theta)$

This is not a specific number — it depends on $\theta$. Unless $\theta$ is specified, we cannot evaluate numerically.

But perhaps this is a typo? Or maybe it's meant to be $\cos^2(0)$?

If it's $\cos^2(0)$: $\cos(0) = 1$, so $\cos^2(0) = 1^2 = \boxed{1}$

But as written: $\cos^2(\theta)$ → depends on $\theta$, so cannot evaluate without knowing $\theta$.

But since it's listed with others, likely a typo. Maybe it's $\cos^2(0)$?

Assume it's $\cos^2(0)$: then answer is $\boxed{1}$

Alternatively, if it's truly $\cos^2(\theta)$, then the expression remains $\boxed{\cos^2(\theta)}$

But probably intended to be $\cos^2(0)$ → $\boxed{1}$

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15. $\tan^{-1}(-\sqrt{3})$

We know: $\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}$, and $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$$
\tan^{-1}(-\sqrt{3}) = \boxed{-\frac{\pi}{3}}
$$

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16. $\arccos(1)$

$\cos(0) = 1$, and $0 \in [0, \pi]$

$$
\arccos(1) = \boxed{0}
$$

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17. $\arccos(0)$

$\cos\left(\frac{\pi}{2}\right) = 0$, and $\frac{\pi}{2} \in [0, \pi]$

$$
\arccos(0) = \boxed{\frac{\pi}{2}}
$$

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18. $\tan^2(0)$

$\tan(0) = 0$, so $\tan^2(0) = 0^2 = \boxed{0}$

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19. $\arcsin(-1)$

$\sin\left(-\frac{\pi}{2}\right) = -1$, and $-\frac{\pi}{2} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

$$
\arcsin(-1) = \boxed{-\frac{\pi}{2}}
$$

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20. $\arccos(-1)$

$\cos(\pi) = -1$, and $\pi \in [0, \pi]$

$$
\arccos(-1) = \boxed{\pi}
$$

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Now, go back to #9: $\arctan\left(-\frac{4}{3}\right)$

As discussed, this is not a standard angle, so we cannot express it as a multiple of $\pi$ without a calculator.

But since the problem asks to evaluate without a calculator, and the others are all standard angles, this might be a typo.

Possibility: Did they mean $\arctan\left(-\frac{\sqrt{3}}{3}\right)$? That would be $-\frac{\pi}{6}$

Or $\arctan(-1)$? Already done.

Alternatively, perhaps it's $\arctan\left(\frac{1}{\sqrt{3}}\right)$? But that's $\frac{\pi}{6}$

Given the fraction $-\frac{4}{3}$, and no standard angle has tangent equal to $\frac{4}{3}$, we must conclude:

$$
\arctan\left(-\frac{4}{3}\right) = \boxed{-\arctan\left(\frac{4}{3}\right)}
$$

But since the problem likely expects a numerical answer in terms of $\pi$, and none exists, it's possible that the problem meant something else.

Alternatively, perhaps it's part of a triangle? But no.

So, assuming the problem is correct, and we are to leave it as is, then:

$$
\boxed{-\arctan\left(\frac{4}{3}\right)}
$$

But if the intention was to have only standard angles, maybe it's a typo.

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Final Answers:

1. $\boxed{\frac{\pi}{6}}$
2. $\boxed{\frac{\pi}{3}}$
3. $\boxed{\frac{\pi}{6}}$
4. $\boxed{\frac{\pi}{6}}$
5. $\boxed{\frac{\pi}{6}}$
6. $\boxed{\frac{\pi}{4}}$
7. $\boxed{-\frac{\pi}{6}}$
8. $\boxed{\frac{2\pi}{3}}$
9. $\boxed{-\arctan\left(\frac{4}{3}\right)}$ (or possibly a typo)
10. $\boxed{\frac{5\pi}{6}}$
11. $\boxed{-\frac{\pi}{4}}$
12. $\boxed{-\frac{\pi}{4}}$
13. $\boxed{0}$
14. $\boxed{\cos^2(\theta)}$ (but likely meant $\cos^2(0) = \boxed{1}$)
15. $\boxed{-\frac{\pi}{3}}$
16. $\boxed{0}$
17. $\boxed{\frac{\pi}{2}}$
18. $\boxed{0}$
19. $\boxed{-\frac{\pi}{2}}$
20. $\boxed{\pi}$

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Let me know if you'd like clarification on any!