Question

trig/algebra problem, please help! :)

Answer 1

Hi, I'm having trouble solving -2sin(t)+2cos(2t)=0

Answer 2

I factored out the -2 so that I ended up with sin(t)-cost(2t)=0, but I'm confused about what I should do to get t by itself.

Answer 3

cost(2t)? there's really t after cos?

Answer 4

Well, here's the problem. I need to find the critical points of f(t)= 2cost(t)+sin(t) on the interval [0,pi/2]. So I found the derivative, which is -2sin(t)+2cost(2t). Then I set that to equal 0 to find the critical points. I factored out a -2 and got sin(t)-cos(2t)=0. Now I just need to solve for t, but I'm confused.

Answer 5

whoops, that should say f(t)-2cost(t)+sin(2t) at the beginning

Answer 6

you're confusing.....

Answer 7

Ah, should probably start over on this one.

Answer 8

How am I confusing? The derivative needs to be set to 0 to find the critical points....

Answer 9

Okay, did I do something wrong?

Answer 10

Just want to be clear on what is going on.
Original function is f(t)=2cos(t)+sin(2t) ?
and you found f'(t) = -2sin(t)+2cos(2t) ?

Answer 11

correct

Answer 12

Ok, use cosine double-angle identity.
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 13

(I recommend the one that only has sine in it)

Answer 14

i was referring to your cosines. sometimes you write cos(2t) sometimes cost(2t)

Answer 15

alright, so then when I simplify the derivative down to sin(t)=cos(2t) I should use the double angle formula to see that that would equal cos(2t)=cos^2(t)-sin^2(t)?

Answer 16

okay, sorry about that Igbasallote

Answer 17

You could use that one, but it would be easier with cos(2t) = 1-2sin^2(t).

Answer 18

Then your equation will only be in terms of sin(t).

Answer 19

So then I set it up to be 1-2sin^2(t)=0? Okay, so that would be sin^2(t)=0.5 right so far?

Answer 20

No, because your equation is sin(t)-cos(2t)=0, after replacing cos(2t) with 1-2sin^2(t), you'll have
sin(t)-1+2sin^2(t)=0

Answer 21

Oh, I see. so far then I have sin(t)(1+2sin(t))=1 is that good so far?

Answer 22

It's correct, but not easy to solve in that form, think quadratic equation.

Answer 23

Okay, so should I replace the sin's with x's to make it 2x^2+x-1=0 ?

Answer 24

That should be mighty easy to solve, yes. :-)

Answer 25

Okay, I should be able to do the rest then. Thank you so much!!!!

Answer 26

Caveat: you'll get two solutions to that quadratic, but remember your domain.

Answer 27

Okay, yeah, one of them's negative. Thanks again :)