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Sketch arc trig

  • Thread starter jwxie
  • Start date
  • #1
280
0

Homework Statement



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

Homework Equations




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?
 

Answers and Replies

  • #2
33,496
5,188

Homework Statement



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

Homework Equations




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?
 
  • #3
280
0
Hi, thank you for pointing it out the trend.

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

What about cases like this
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 ?
 
  • #4
33,496
5,188
Hi, thank you for pointing it out the trend.

So in general, I should always get x by itself?
Generally, yes.
What about cases like this
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 ?
Take it all the way to solve for x.

[tex]x = \pm \sqrt{\frac{1}{1 - sin(y)}}[/tex]

There should be some restrictions on both x and y so that the +/- in the equation above aren't needed.
 

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