SOLUTION: I have a similar question to one that has been answered.
The vertical displacement of the end of a robot arm (in cm) at time t (in seconds) is given by y = 8+7cos3t + 7cos6t
1.
Algebra.Com
Question 966917: I have a similar question to one that has been answered.
The vertical displacement of the end of a robot arm (in cm) at time t (in seconds) is given by y = 8+7cos3t + 7cos6t
1. Find all times, t>0 (fractions of pi) where the vertical displacement is 8cm.
2. What is the first time t>0 that the vertical diplacement is 8cm?
This is what I have started doing:
Let x = 3t
y = 8 + 7cos(x) + 7cos(2x)
now do I divide out 7 (from cos(x) and cos(2x)) and use a trigonometric Identity cos(2x)? which gives me cos(x) + 2cos^2x-1
Any help in the right direction would be appreciated. Thankyou
Chad.
Answer by Murthy(22) (Show Source): You can put this solution on YOUR website!
Given, y = 8+7cos3t + 7cos6t
when y=8 cm
8=8+7cos3t + 7cos6t
can be rearranged as
7(cos3t+cos6t)=0
or cos3t+cos6t =0
2cos(9t/2).cos (3t/2)=0 {cosA + cosB = 2 cos(A+B/2).cos(A-B/2)}
or cos(9t/2).cos (3t/2)=0
This means either cos(9t/2)=0 or cos (3t/2)=0
if cos(9t/2)=0
9t/2= n(pi)/2
t=n(pi)/9
if cos (3t/2)=0
3t/2= n(pi)/2
t=n(pi)/3
The first time when displacement is 8 cm is t=pi/9
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