Simplify Inverse Trigonometric Functions

Simplify the expressions:

1. $\sin(\arcsin(x))$ and $\arcsin(\sin(x))$
2. $\cos(\arccos(x))$ and $\arccos(\cos(x))$
3. $\tan(\arctan(x))$ and $\arctan(\tan(x))$

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Solution:

1. Sine and arcsine are inverses of each other. Therefore, the properties of inverse functions may be used to write:

   $$ \sin(\arcsin(x)) = x, \quad \text{for } -1 \leq x \leq 1 $$

   $$ \arcsin(\sin(x)) = x, \quad \text{for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$

   NOTE: If $x$ in $\arcsin(\sin(x))$ is not in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, find $\theta$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin(x) = \sin(\theta)$ and then simplify: $\arcsin(\sin(x)) = \theta$

2. Cosine and arccosine are inverse functions of each other:

   $$ \cos(\arccos(x)) = x, \quad \text{for } -1 \leq x \leq 1 $$

   $$ \arccos(\cos(x)) = x, \quad \text{for } x \in [0, \pi] $$

   NOTE: If $x$ in $\arccos(\cos(x))$ is not in the interval $[0, \pi]$, find $\theta$ in the interval $[0, \pi]$ such that $\cos(x) = \cos(\theta)$, and then simplify $\arccos(\cos(x)) = \theta$.

3. Tangent and arctangent are inverses of each other:

   $$ \tan(\arctan(x)) = x $$

   $$ \arctan(\tan(x)) = x \quad \text{for } x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $$

   NOTE: If $x$ in $\arctan(\tan(x))$ is not in the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, find $\theta$ in the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ such that $\tan(x) = \tan(\theta)$ and then simplify $\arctan(\tan(x)) = \theta$

Express the following as algebraic expressions:

$$ \sin(\arccos(x)) \quad \text{and} \quad \tan(\arccos(x)) $$

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Solution:

Let $A = \arccos(x)$. Hence,

$$ \cos(A) = \cos(\arccos(x)) = x $$

Use a right triangle with angle $A$ such that $\cos(A) = x$ (or $\frac{x}{1}$). Find the second leg and calculate $\sin(A)$ and $\tan(A)$.

$$ \sin(\arccos(x)) = \sin(A) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} \quad \text{for } x \in [-1, 1] $$

$$ \tan(\arccos(x)) = \tan(A) = \frac{\sqrt{1 - x^2}}{x} \quad \text{for } x \in [-1, 0) \cup (0, 1] $$

Express the following as algebraic expressions:

$$ \cos(\arcsin(x)) \quad \text{and} \quad \tan(\arcsin(x)) $$

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Solution:

Let $A = \arcsin(x)$. Hence,

$$ \sin(A) = \sin(\arcsin(x)) = x $$

Use a right triangle with angle $A$ such that $\sin(A) = x$ (or $\frac{x}{1}$), find the second leg and calculate $\cos(A)$ and $\tan(A)$.

$$ \cos(\arcsin(x)) = \cos(A) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} \quad \text{for } x \in [-1, 1] $$

$$ \tan(\arcsin(x)) = \tan(A) = \frac{x}{\sqrt{1 - x^2}} \quad \text{for } x \in (-1, 1) $$

Express the following as algebraic expressions:

$$ \sin(\arctan(x)) \quad \text{and} \quad \cos(\arctan(x)) $$

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Solution:

Let $A = \arctan(x)$. Hence,

$$ \tan(A) = \tan(\arctan(x)) = x $$

Use a right triangle with angle $A$ such that $\tan(A) = x$ (or $\frac{x}{1}$). Find the hypotenuse and calculate $\sin(A)$ and $\cos(A)$.

$$ \sin(\arctan(x)) = \sin(A) = \frac{x}{\sqrt{1 + x^2}} $$

$$ \cos(\arctan(x)) = \cos(A) = \frac{1}{\sqrt{1 + x^2}} $$

Simplify the following expressions:

1. $\arccos(0)$, $\arcsin(-1)$, $\arctan(-1)$
2. $\sin(\arcsin(-1/2))$, $\arccos(\cos(\pi/2))$, $\arccos(\cos(-\pi/2))$
3. $\cos(\arcsin(-1/2))$, $\arcsin(\sin(\pi/3))$, $\arcsin(\tan(3\pi/4))$
4. $\arccos(\tan(7\pi/4))$, $\arcsin(\sin(13\pi/3))$, $\arctan(\tan(-17\pi/4))$, $\arcsin(\sin(9\pi/5))$

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Solution:

1. Use the direct definitions of inverse functions:

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

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

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

2. Simplify inner functions then the outer functions using definitions:

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

   $$ \arccos\left(\cos\left(\frac{\pi}{2}\right)\right) = \arccos(0) = \frac{\pi}{2} $$

   $$ \arccos\left(\cos\left(-\frac{\pi}{2}\right)\right) = \arccos(0) = \frac{\pi}{2} $$

3. Simplify inner functions then the outer functions:

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

   $$ \arcsin\left(\sin\left(\frac{\pi}{3}\right)\right) = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} $$

   $$ \arcsin\left(\tan\left(\frac{3\pi}{4}\right)\right) = \arcsin(-1) = -\frac{\pi}{2} $$

4. Adjust the angles to fall within the correct intervals before applying inverse functions:

   $$ \arccos\big(\tan(7\pi/4)\big) = \arccos(-1) = \pi $$

   $$ \arcsin\big(\sin(13\pi/3)\big) = \arcsin\big(\sin(4\pi + \pi/3)\big) = \arcsin\big(\sin(\pi/3)\big) = \pi/3 $$

   $$ \arctan\big(\tan(-17\pi/4)\big) = \arctan\big(\tan(-4\pi - \pi/4)\big) = \arctan\big(\tan(-\pi/4)\big) = -\pi/4 $$

   $$ \arcsin\big(\sin(9\pi/5)\big) = \arcsin\big(\sin(2\pi - \pi/5)\big) = \arcsin\big(\sin(-\pi/5)\big) = -\pi/5 $$

Let $A = \arcsin(2/3)$ and $B = \arccos(-1/2)$. Find the exact value of $\sin(A + B)$.

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Solution:

Use the angle addition identity:

$$ \sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) $$

Expand the given expression by substituting $A$ and $B$:

$$ \sin(A + B) = \sin(\arcsin(\tfrac{2}{3}))\cos(\arccos(-\tfrac{1}{2})) + \cos(\arcsin(\tfrac{2}{3}))\sin(\arccos(-\tfrac{1}{2})) $$

Use inverse identities to simplify each individual term:

• $\sin(\arcsin(\tfrac{2}{3})) = \tfrac{2}{3}$
• $\cos(\arccos(-\tfrac{1}{2})) = -\tfrac{1}{2}$
• $\cos(\arcsin(\tfrac{2}{3})) = \sqrt{1 - \left(\tfrac{2}{3}\right)^2} = \sqrt{\tfrac{5}{9}} = \tfrac{\sqrt{5}}{3}$
• $\sin(\arccos(-\tfrac{1}{2})) = \sqrt{1 - \left(-\tfrac{1}{2}\right)^2} = \sqrt{\tfrac{3}{4}} = \tfrac{\sqrt{3}}{2}$

Substitute these values back into the expanded equation and calculate:

$$ \sin(A + B) = \left(\tfrac{2}{3}\right)\left(-\tfrac{1}{2}\right) + \left(\tfrac{\sqrt{5}}{3}\right)\left(\tfrac{\sqrt{3}}{2}\right) = -\tfrac{1}{3} + \tfrac{\sqrt{15}}{6} $$

Write $Y = \sin(2 \arcsin(x))$ as an algebraic expression.

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Solution:

Let $A = \arcsin(x)$. Hence $Y$ may be written as:

$$ Y = \sin(2A) $$

Use the double angle identity $\sin(2A) = 2 \sin(A) \cos(A)$ to rewrite $Y$:

$$ Y = 2 \sin(A) \cos(A) = 2 \sin(\arcsin(x)) \cos(\arcsin(x)) $$

Apply the standard identities $\sin(\arcsin(x)) = x$ and $\cos(\arcsin(x)) = \sqrt{1 - x^2}$ to yield the final expression:

$$ Y = 2 x \sqrt{1 - x^2} $$

Find the exact value of $Y = \sin(2 \arctan(3/4))$.

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Solution:

Let $A = \arctan\left(\frac{3}{4}\right)$. Hence $Y$ may be written as:

$$ Y = \sin(2A) = 2 \sin(A) \cos(A) $$

Evaluate the inner functions using the standard algebraic conversion properties for arctangent:

$$ \sin(A) = \sin\left(\arctan\left(\frac{3}{4}\right)\right) = \frac{\frac{3}{4}}{\sqrt{1 + \left(\frac{3}{4}\right)^2}} = \frac{3}{5} $$

$$ \cos(A) = \cos\left(\arctan\left(\frac{3}{4}\right)\right) = \frac{1}{\sqrt{1 + \left(\frac{3}{4}\right)^2}} = \frac{4}{5} $$

Substitute these into the double angle formula:

$$ Y = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25} $$