Math / Trigonometry

QuestionProb. 6 Compute the following exactly, without using a calculator. (a) $\arcsin \left(\frac{\sqrt{3}}{2}\right)$ (b) $\arccos \left(-\frac{1}{2}\right)$ (c) $\arctan (-1)$ (d) $\arccos \left(\cos \left(\frac{\pi}{3}\right)\right)$ (e) $\arcsin \left(\sin \left(\frac{2 \pi}{3}\right)\right)$ (Be careful on this one; the answer is not $\frac{2 \pi}{3}!$ ) (f) $\arctan (0.81)$ (Okay, you can use your calculator on this one!)

Studdy Solution

STEP 1

1. We are working with inverse trigonometric functions: arcsin, arccos, and arctan.
2. We need to find exact values for parts (a) through (e) without using a calculator.
3. For part (f), we are allowed to use a calculator to approximate the value.

STEP 2

1. Solve for $\arcsin \left(\frac{\sqrt{3}}{2}\right)$ .
2. Solve for $\arccos \left(-\frac{1}{2}\right)$ .
3. Solve for $\arctan (-1)$ .
4. Solve for $\arccos \left(\cos \left(\frac{\pi}{3}\right)\right)$ .
5. Solve for $\arcsin \left(\sin \left(\frac{2 \pi}{3}\right)\right)$ .
6. Approximate $\arctan (0.81)$ using a calculator.

STEP 3

Recall that $\arcsin$ gives the angle whose sine is the given value. We know that:
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefore:
$\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

STEP 4

Recall that $\arccos$ gives the angle whose cosine is the given value. We know that:
$\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$
Therefore:
$\arccos \left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

STEP 5

Recall that $\arctan$ gives the angle whose tangent is the given value. We know that:
$\tan \left(-\frac{\pi}{4}\right) = -1$
Therefore:
$\arctan (-1) = -\frac{\pi}{4}$

STEP 6

The $\arccos$ function will return the principal value, which is the angle itself since $\cos$ is a one-to-one function in the range $[0, \pi]$ . Therefore:
$\arccos \left(\cos \left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$

STEP 7

The $\arcsin$ function will return the principal value, which must be in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$ . We know:
$\sin \left(\frac{2\pi}{3}\right) = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
However, the principal value for $\arcsin\left(\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi}{3}$ , not $\frac{2\pi}{3}$ . Therefore:
$\arcsin \left(\sin \left(\frac{2\pi}{3}\right)\right) = \frac{\pi}{3}$

STEP 8

Use a calculator to approximate:
$\arctan (0.81) \approx 0.68 \, \text{radians}$
The solutions are: (a) $\frac{\pi}{3}$ (b) $\frac{2\pi}{3}$ (c) $-\frac{\pi}{4}$ (d) $\frac{\pi}{3}$ (e) $\frac{\pi}{3}$ (f) $\approx 0.68$

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Studdy Solution

STEP 1

1. We are working with inverse trigonometric functions: arcsin, arccos, and arctan.
2. We need to find exact values for parts (a) through (e) without using a calculator.
3. For part (f), we are allowed to use a calculator to approximate the value.

STEP 2

STEP 3

Recall that $\arcsin$ gives the angle whose sine is the given value. We know that:
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefore:
$\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

STEP 4

Recall that $\arccos$ gives the angle whose cosine is the given value. We know that:
$\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$
Therefore:
$\arccos \left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

STEP 5

Recall that $\arctan$ gives the angle whose tangent is the given value. We know that:
$\tan \left(-\frac{\pi}{4}\right) = -1$
Therefore:
$\arctan (-1) = -\frac{\pi}{4}$

STEP 6

The $\arccos$ function will return the principal value, which is the angle itself since $\cos$ is a one-to-one function in the range $[0, \pi]$ . Therefore:
$\arccos \left(\cos \left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$

STEP 7

STEP 8

Use a calculator to approximate:
$\arctan (0.81) \approx 0.68 \, \text{radians}$
The solutions are: (a) $\frac{\pi}{3}$ (b) $\frac{2\pi}{3}$ (c) $-\frac{\pi}{4}$ (d) $\frac{\pi}{3}$ (e) $\frac{\pi}{3}$ (f) $\approx 0.68$

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Studdy Solution

STEP 1

1. We are working with inverse trigonometric functions: arcsin, arccos, and arctan.2. We need to find exact values for parts (a) through (e) without using a calculator.3. For part (f), we are allowed to use a calculator to approximate the value.

STEP 2

STEP 3

Recall that arcsin⁡\arcsinarcsin gives the angle whose sine is the given value. We know that:sin⁡(π3)=32 \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} sin(3π​)=23​​Therefore:arcsin⁡(32)=π3 \arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} arcsin(23​​)=3π​

STEP 4

Recall that arccos⁡\arccosarccos gives the angle whose cosine is the given value. We know that:cos⁡(2π3)=−12 \cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2} cos(32π​)=−21​Therefore:arccos⁡(−12)=2π3 \arccos \left(-\frac{1}{2}\right) = \frac{2\pi}{3} arccos(−21​)=32π​

STEP 5

Recall that arctan⁡\arctanarctan gives the angle whose tangent is the given value. We know that:tan⁡(−π4)=−1 \tan \left(-\frac{\pi}{4}\right) = -1 tan(−4π​)=−1Therefore:arctan⁡(−1)=−π4 \arctan (-1) = -\frac{\pi}{4} arctan(−1)=−4π​

STEP 6

STEP 7

STEP 8

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1. We are working with inverse trigonometric functions: arcsin, arccos, and arctan.
2. We need to find exact values for parts (a) through (e) without using a calculator.
3. For part (f), we are allowed to use a calculator to approximate the value.

Recall that $\arcsin$ gives the angle whose sine is the given value. We know that:
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefore:
$\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

Recall that $\arccos$ gives the angle whose cosine is the given value. We know that:
$\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$
Therefore:
$\arccos \left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

Recall that $\arctan$ gives the angle whose tangent is the given value. We know that:
$\tan \left(-\frac{\pi}{4}\right) = -1$
Therefore:
$\arctan (-1) = -\frac{\pi}{4}$