+0  
 
+2
127
10
avatar+163 

Consider the following points on the graphs of \(y = \cos{x}\) and \(y = \cos{\frac{x}{2}}\) with the added vertical dashed lines:


Find C and D

I tried finding the cosine of points a and b but not sure how I continue here

 Jan 30, 2023
edited by YourAverageDummy  Jan 30, 2023
 #5
avatar+23 
+1

B= 4/8 C=4/8

 Jan 31, 2023
 #6
avatar+23 
0

Sorry C is -1/8

Ds1234  Jan 31, 2023
 #7
avatar+163 
0

no but I would be interested how you came to that solution

YourAverageDummy  Jan 31, 2023
 #8
avatar+118573 
+1

I think there are a couple of different ways to take lthis question.

 

Looking at the graph I can see that  b=a+2pi

From a to b is a whole wavelength of  cos a   and a half a wavelength of cos (a/2)

This means that

1-d = c - - 1

1-d= c+1

c=-d

 

\(cos\;a= \frac{1}{8}\\ cos(\frac{a}{2}+\frac{a}{2})=\frac{1}{8}\\ cos^2(\frac{a}{2})-sin^2(\frac{a}{2})=\frac{1}{8}\\ cos^2(\frac{a}{2})-(1-cos^2(\frac{a}{2}))=\frac{1}{8}\\ 2cos^2(\frac{a}{2})-1=\frac{1}{8}\\ 2cos^2(\frac{a}{2})=\frac{9}{8}\\ cos^2(\frac{a}{2})=\frac{9}{16}\\ cos(\frac{a}{2})=\pm\frac{3}{4}\\ \text{discount the positive answer}\\ c=cos(\frac{a}{2})=-\frac{3}{4}\\ \)

 

\(d=\frac{3}{4}\)

 Jan 31, 2023
edited by Melody  Jan 31, 2023
 #9
avatar+163 
+1

huh. I never thought that you could prove c in terms of d
Thanks Melody!

 Jan 31, 2023
 #10
avatar+118573 
0

You are welcome :)

Melody  Feb 1, 2023

2 Online Users