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  2. AP Pre Calculus
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Why are domain restrictions necessary for inverse trig functions?

Trig functions are periodic, so domain restrictions ensure unique inverse values.

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Why are domain restrictions necessary for inverse trig functions?

Trig functions are periodic, so domain restrictions ensure unique inverse values.

What is the range of arcsine?

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}][−2π​,2π​]

What is the range of arccosine?

[0,π][0, \pi][0,π]

What is the range of arctangent?

(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})(−2π​,2π​)

Explain how the unit circle helps evaluate inverse trig functions.

It visually represents angles and their corresponding sine, cosine, and tangent values, making it easy to find the angle for a given ratio.

Describe the relationship between trig functions and inverse trig functions.

Inverse trig functions 'undo' what trig functions do. If sin⁡(x)=y\sin(x) = ysin(x)=y, then arcsin⁡(y)=x\arcsin(y) = xarcsin(y)=x.

How do you evaluate arcsin⁡(12)\arcsin(\frac{1}{2})arcsin(21​)?

Find the angle within the range of arcsine ([−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}][−2π​,2π​]) whose sine is 12\frac{1}{2}21​. Answer: π6\frac{\pi}{6}6π​.

How do you evaluate arccos⁡(0)\arccos(0)arccos(0)?

Find the angle within the range of arccosine ([0,π][0, \pi][0,π]) whose cosine is 0. Answer: π2\frac{\pi}{2}2π​.

How do you evaluate arctan⁡(1)\arctan(1)arctan(1)?

Find the angle within the range of arctangent (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})(−2π​,2π​) whose tangent is 1. Answer: π4\frac{\pi}{4}4π​.

How to solve sin⁡(arccos⁡(x))\sin(\arccos(x))sin(arccos(x))?

Let θ=arccos⁡(x)\theta = \arccos(x)θ=arccos(x). Then cos⁡(θ)=x\cos(\theta) = xcos(θ)=x. Draw a right triangle, find the missing side using the Pythagorean theorem, and then find sin⁡(θ)\sin(\theta)sin(θ).

Evaluate tan⁡(arcsin⁡(35))\tan(\arcsin(\frac{3}{5}))tan(arcsin(53​))

Let θ=arcsin⁡(35)\theta = \arcsin(\frac{3}{5})θ=arcsin(53​). Then sin⁡(θ)=35\sin(\theta) = \frac{3}{5}sin(θ)=53​. Draw a right triangle, find the adjacent side using the Pythagorean theorem (which is 4), and then find tan⁡(θ)=34\tan(\theta) = \frac{3}{4}tan(θ)=43​.

Evaluate cos⁡(arctan⁡(2))\cos(\arctan(2))cos(arctan(2))

Let θ=arctan⁡(2)\theta = \arctan(2)θ=arctan(2). Then tan⁡(θ)=2\tan(\theta) = 2tan(θ)=2. Draw a right triangle, find the hypotenuse using the Pythagorean theorem (which is 5\sqrt{5}5​), and then find cos⁡(θ)=15\cos(\theta) = \frac{1}{\sqrt{5}}cos(θ)=5​1​.

Describe the key features of the graph of y=arcsin⁡(x)y = \arcsin(x)y=arcsin(x).

Domain: [-1, 1], Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}][−2π​,2π​], increasing function, symmetric about the origin.

Describe the key features of the graph of y=arccos⁡(x)y = \arccos(x)y=arccos(x).

Domain: [-1, 1], Range: [0,π][0, \pi][0,π], decreasing function, no symmetry.

Describe the key features of the graph of y=arctan⁡(x)y = \arctan(x)y=arctan(x).

Domain: (−∞,∞)(-\infty, \infty)(−∞,∞), Range: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})(−2π​,2π​), increasing function, symmetric about the origin, horizontal asymptotes at y=±π2y = \pm \frac{\pi}{2}y=±2π​.

How does the range restriction of arcsin affect its graph?

The graph is bounded between y=−π2y = -\frac{\pi}{2}y=−2π​ and y=π2y = \frac{\pi}{2}y=2π​, ensuring it's a function (passes the vertical line test).

How does the range restriction of arccos affect its graph?

The graph is bounded between y=0y = 0y=0 and y=πy = \piy=π, ensuring it's a function (passes the vertical line test).

How does the range restriction of arctan affect its graph?

The graph is bounded between y=−π2y = -\frac{\pi}{2}y=−2π​ and y=π2y = \frac{\pi}{2}y=2π​, ensuring it's a function (passes the vertical line test).