ONE-EIGHTH STAGE, ROUND 2 SPEED RACE
2021
St. James Seminary SHS: 37 points
Berekum SHS: 32 points
Samuel Otu Presby SHS: 23 points
QUESTION
Find the inverse of $y=\sin\left(\dfrac{x}{2}\right)$ for $-\pi < x < \pi$.
ANSWER: $y=2\sin^{-1}(x)$
SOLUTION
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ for $-\pi < x < \pi$.
is written as $y=a\sin^{-1}x$.
ANSWER: $y=2\sin^{-1}(x)$
SOLUTION
Swap the variables $x$ and $y$ to begin finding the inverse function:
$x = \sin\left(\dfrac{y}{2}\right)$
Take the inverse sine ($\sin^{-1}$) of both sides to isolate the angle expression:
$\sin^{-1}(x) = \dfrac{y}{2}$
Multiply both sides by 2 to solve explicitly for $y$:
$y = 2\sin^{-1}(x)$
PRACTICE QUESTIONS
1. Find the inverse of $y=\sin\left(\dfrac{x}{2}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=2\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=2$, so:
$y=2\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{2}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{2}$
Multiply both sides by $2$:
$y=2\sin^{-1}(x)$
2. Find the inverse of $y=\sin\left(\dfrac{x}{3}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=3\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=3$, so:
$y=3\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{3}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{3}$
Multiply both sides by $3$:
$y=3\sin^{-1}(x)$
3. Find the inverse of $y=\sin\left(\dfrac{x}{4}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=4\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=4$, so:
$y=4\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{4}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{4}$
Multiply both sides by $4$:
$y=4\sin^{-1}(x)$
4. Find the inverse of $y=\sin\left(\dfrac{x}{5}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=5\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=5$, so:
$y=5\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{5}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{5}$
Multiply both sides by $5$:
$y=5\sin^{-1}(x)$
5. Find the inverse of $y=\sin\left(\dfrac{x}{6}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=6\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=6$, so:
$y=6\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{6}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{6}$
Multiply both sides by $6$:
$y=6\sin^{-1}(x)$
6. Find the inverse of $y=\sin\left(\dfrac{x}{8}\right)$ for $-\pi\lt x\lt\pi$.
ANSWER: $y=8\sin^{-1}(x)$
SOLUTION 1
The inverse of any function of the form $y=\sin\left(\dfrac{x}{a}\right)$ is $y=a\sin^{-1}x$.
Here $a=8$, so:
$y=8\sin^{-1}(x)$
SOLUTION 2
Swap the variables $x$ and $y$:
$x=\sin\left(\dfrac{y}{8}\right)$
Take the inverse sine of both sides:
$\sin^{-1}(x)=\dfrac{y}{8}$
Multiply both sides by $8$:
$y=8\sin^{-1}(x)$