# Sketch arc trig

## Homework Statement

Sketch y = cos-1(2x) and y = sin-1(3x)

## The Attempt at a Solution

This is my attempt

cos-1(2x) = y, which means cos y = 2x

produce a table with y from (-pi/2 to pi/2)

What I want to confirm is the 2x. I need do 1/2 x instead of 2x when we calculate the x?
And 1/3x for the other problem?

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Mark44
Mentor

## Homework Statement

Sketch y = cos-1(2x) and y = sin-1(3x)

## The Attempt at a Solution

This is my attempt

cos-1(2x) = y, which means cos y = 2x
so (1/2) cos y = x
produce a table with y from (-pi/2 to pi/2)
No, y is in [0, pi]. That's the vertical axis. If you plot pairs of points (x, y), you will have the graph of x = (1/2)cos(y), which is also the graph of y = cos-1(2x).
What I want to confirm is the 2x. I need do 1/2 x instead of 2x when we calculate the x?
And 1/3x for the other problem?

Hi, thank you for pointing it out the trend.

So in general, I should always get x by itself?

sin-1(1-1/x^2), which gives siny = 1-1/x^2
do i solve it as 1-siny = 1/x^2, 1-siny = (1/x)^2, sqrt(1-siny) = 1/x ?

Mark44
Mentor
Hi, thank you for pointing it out the trend.

So in general, I should always get x by itself?
Generally, yes.
$$x = \pm \sqrt{\frac{1}{1 - sin(y)}}$$