trigonometric functions

Question

Hi,

 

Am having trouble with this question and any help would be greatly appreciated.

 

The heights of two different robot arms (in cm) at time t (in seconds) are given by

 

y1 =6+2sin6t+4sin3t,     and

 

y2 =8+2cos3t.

 

(a) Find all times, t > 0, (in exact form i.e. in terms of fractions of π) when the height of both robot arms are the same, i.e. y1 = y2.

 

(b) What is the first time, t > 0, that the heights of both robot arms are the same? Give your answer exactly and to 3 decimal places.

 

 

Thanks in advance

 

Michael

 

Answer 1

Hi Michael!

 

If we want y1=y2, then we need to set the two equations equal to each other and try to solve for t:

6+2sin6t+4sin3t=8+2cos3t

 

First, I would subtract 6 from both sides to get all our constants together:

 

2sin6t+4sin3t=2+2cos3t

 

With these types of trig equations, it's usually easiest to solve if all of the trig functions have the same argument (i.e. the same thing plugged into all of them), so we should try to change that sin6t. If we think of sin6t=sin(2*3t) then we can use a double angle formula: sin(2x)=2sinx cosx. So, your equation becomes:

 

2(2sin3t cos3t)+4sin3t=2(1+cos3t), so

4sin3t cos3t+4sin3t=2(1+cos3t)

 

On the left hand side, we can factor a 4sin3t out of everything to get:

4sin3t (cos3t+1)=2(1+cos3t). Subtracting to get everything on the same side of the equal sign gives us

 

4sin3t(cos3t+1)-2(1+cos3t)=0. Now we can factor a (1+cos3t) out of the left hand side to be left with

 

(1+cos3t)(4sin3t-2)=0.

 

And now we're in business, because we have two things times each other equalling 0, so we can set them equal to zero separately, which is a lot easier!

 

First, we have to solve 1+cos3t=0, which is the same as cos3t=-1. From the unit circle, we know that this happens when 3t=π (pi). But that's not the only place! It could also be 3t=3π, or 3t=5π,.... So there are infinitely many solutions. The best way to write this is:

3t=π+2πk, where k=0,1,2,..., so if we divide both sides by 3, we get

t=(1/3)(π+2πk), where k=0,1,2,...

 

That gives us half of our answer, and now we need to set the other part of the equation equal to zero and solve:

4sin3t-2=0, so

4sin3t=2, or

sin3t=1/2.

 

Again, from the unit circle, this means either 3t=π/6 or 3t=5π/6, and multiples of 2π after those like in the previous solution. I'll let you try to finish the problem from here. Please let me know if you have any more questions! (We can also schedule a private one on one tutoring session if you would like further help).

 

Erika