My Math Forum Transforming a graph cos(x) to cos(2x-60) in correct order?

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 April 25th, 2016, 08:36 AM #1 Member   Joined: Sep 2014 From: UK Posts: 82 Thanks: 1 Transforming a graph cos(x) to cos(2x-60) in correct order? So I was thinking cos(x) --> cos(2x) --> cos(2x-60) is the correct order, since cos(x) --> cos(x-60) --> cos(2(x-60)) seems to me what would happen if I done the translation of 60 in x and then a stretch of 1/2 parallel to x. But the book gives me the order of the translation and then the stretch, and says it's cos(2x-60). Does the order matter in this case? Last edited by skipjack; April 25th, 2016 at 11:15 AM.
 April 25th, 2016, 08:58 AM #2 Senior Member   Joined: Dec 2015 From: holland Posts: 163 Thanks: 37 Math Focus: tetration Yes 2x - 60 is not equal to 2(x - 60).
April 25th, 2016, 10:24 AM   #3
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Quote:
 Originally Posted by manus Yes 2x - 60 is not equal to 2(x - 60).
So would you agree that the stretch of 1/2 parallel to x of y = cos(x-60) would give y = cos(2(x-60)) = cos(2x-120) and so would not equal to y = cos(2x-60), like written in the book?

 April 25th, 2016, 10:25 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,236 Thanks: 2412 Math Focus: Mainly analysis and algebra $\cos x \longrightarrow \cos ( x -30) \longrightarrow \cos 2(x-30) = \cos (2x-60)$ There is no "correct order". Thanks from Tangeton
April 25th, 2016, 10:48 AM   #5
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 Originally Posted by v8archie $\cos x \longrightarrow \cos ( x -30) \longrightarrow \cos 2(x-30) = \cos (2x-60)$ There is no "correct order".
I get what you said there and this is exactly what I thought, that the stretch needs to be applied also to the previous translation.

The reason why I asked for 'correct order' is because the two translations in the book are (60 0) vector and 1/2 scale factor stretch parallel to the x-axis. There is no (30 0).

Quote from the book:
''The curve of y = cos(x-60) is obtained from that of y = cos(x) by a translation of (60 0).
The curve of y = cos(2x-60) is obtained from that of y =cos(x-60) by a stretch of scale factor 1/2 parallel to the x-axis.''

It explicitly stated that it's getting the y = cos(2x-60) from y = cos(x-60) by stretch 1/2 parallel to x, which is not possible.

Having an example from this book that is completely wrong and then beginning to doubt everything I learned is the worst, and I need to confirm that it's wrong with others because I can't trust myself. *Sigh*

Last edited by skipjack; April 25th, 2016 at 11:10 AM.

 April 25th, 2016, 11:14 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,236 Thanks: 2412 Math Focus: Mainly analysis and algebra Yes, the book is wrong.
 April 25th, 2016, 11:20 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,715 Thanks: 1532 No, the book is correct.
 April 25th, 2016, 11:36 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,236 Thanks: 2412 Math Focus: Mainly analysis and algebra The book could be correct if it is determined that we stretch from $x=0$ rather than anywhere else. Last edited by v8archie; April 25th, 2016 at 11:52 AM.
 April 25th, 2016, 11:45 AM #9 Member   Joined: Sep 2014 From: UK Posts: 82 Thanks: 1 After a bit of research I found that the book apparently IS correct. cos(x) --> cos(x-60) --> cos(2x-60). Because in fact cos(x) --> cos(2x) --> cos(2(x-60)) ...I'll just take it for a fact o_o
 April 25th, 2016, 12:14 PM #10 Global Moderator   Joined: Dec 2006 Posts: 18,715 Thanks: 1532 The book is correct. It probably explains somewhere that the assumption v8archie refers to is $x=0$.

 Tags correct, cos2x60, cosx, graph, order, transforming

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