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MelindaR
Group Title
Can somebody help me solve this one?
sinx*cos³x + sin³x*cosx = 1/2
I've got:
cos²x(sinx*cosx) + sin²x(sinx*cosx) = 1/2
(sin²x + cos²x)(sinx*cosx) = 1/2
Since sin²x + cos²x = 1
sinx*cosx = 1/2
How should I proceed now?
 2 years ago
 2 years ago
MelindaR Group Title
Can somebody help me solve this one? sinx*cos³x + sin³x*cosx = 1/2 I've got: cos²x(sinx*cosx) + sin²x(sinx*cosx) = 1/2 (sin²x + cos²x)(sinx*cosx) = 1/2 Since sin²x + cos²x = 1 sinx*cosx = 1/2 How should I proceed now?
 2 years ago
 2 years ago

This Question is Closed

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
P.S. \[0 \le x < 2\]
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Use identity of 2sinxcosx = sin2x
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
I tried, but then I got sin 2x = 1/2, so what?
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
I mean sin2x = 1
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
First, you would get sin2x =1, not 1/2. Agreed?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Ok, so you know where sin m = 1, and that's at pi/2 + 2kpi. Good so far?
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Okay but keep in mind that \[0 \le x < 2\pi\]
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
np. Your FIRST value is pi/4 because sin m = sin 2x where m = pi/2 so x = pi/4 You SECOND and only other value is at x = (5/4)pi because when that is doubled, you are right back at pi/2
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
So, those are the only 2 values for x.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Hmmm, let me see if I get this
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Go ahead, ask questions. I'll try to clarify, but it looks like you already understand most of this.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
I don't understand why x = (5/4)pi is also a solution
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
3/2*pi should also be a value for m, right?
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Hmmm nevermind my last question
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
np. After I show you, I think you'll realize that you already knew the answer, which is good, because you'll get the answers on your own then. If you take (5/4)pi as "x" and multiply by 2, you will get (5/2)pi, which is 2pi + pi/2, so you "cycled" back to pi/2 by going around the circle once and then went pi/2. (3/2)pi will not be an answer because doubled it's 3pi which is 180 degrees and you have sin 180 = 0.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
And how did you find out 5/4pi is an answer? You just guessed?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
The trick is why did I work "backwards" and select (5/4)pi and then double it to "magically" arrive at pi/2, going once around the circle. If that's how I worked it, THAT would be confusing, so I'll explain how I did it. Sometimes you can work "backwards" to get an intuitive idea of where to go, but that doesn't give you the "how to get the answer". You can be intuitive and work backward to "see what you want" and what is really there at the same time. You then have to work forward, the logical way, so that you can "prove" what needs to be proved. So here goes...
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
We know that we have the constraint of 0 <= x <= 2pi. Good. However, our final equation is sin2x = 1. The left side 2x tells us that as x goes from 0 to 2pi, then 2x goes from 0 to 4 pi. Lightbulb time! We go around the circle TWICE! That's the key! So, before resloving x and 2x, we know we have sin (something) = 1 at pi/2 and 5pi/2 because we supposed to twice around. Again, that's the key. Twice around. So, look for halves of those values because x is doubled. So you start intuitively, but back it all up logically. Ask more questions.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Hmmm but actually the constraint is 0<= x < 2pi
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
That's fine. You'll notice that x = pi/4 and x = 5pi/4 and both of those ARE less than 2pi. It doesn't matter that 2x = 5pi/2 > 2pi. The restriction is on x which we are working with. The restriction is not on 2x.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
The "restriction" for 2x is that 2x has to be less than 4pi, not 2pi.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Believe me, I can understand how this could be a little disturbing at first. Keep going over it, and you will see it. You are not expected to see it immediately, but I'm positive about this. If it helps, think of 2x as another variable of think of 2x as a f(x). Or think "twice around" for 2x is "once around" for x.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Hmmm I think I get it. But how would I go about writing that? I left at sin 2x = 1. So let 2x = m and we have m = pi/2. Therefore, 2x = pi/2 and x = pi/4. What about explaining how I got 5/4pi and 5/2pi? I'd need to prove that in an exam, right?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
The way I'd think about it if I put myself in the position of seeing it for the first time would be to say 1) where is sin m = 1? It's at pi/2 and all 2kpi additions to that. 2) I'd know to concentrate on pi/2 right off, but what about those other potential values? Are they "within range". 3) You'd have to relate 0 <= x <= 2pi and 0 <= 2x <= 4pi because that is like a function with y = 2x. You get to use all legitimate values of "Y". And you HAVE to or else you'll get only half credit for your answer. Think of how the sine function is periodic and repeats. That's your conceptual ace in the hole.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Another help: your first value was sin pi/2 = 1, but that's sin 2(pi/4) = 1 for your "x" value because you are working with 2x, so that is your intuitive inclination to consider other values. You had to double something a lot smaller.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
You have to mull things like this over. But don't worry, they expand your mind, they really do. Just be comfortable with it and then it flows.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Wow, you're a great teacher. I get it now, thanks! I always have a little trouble considering values that aren't so apparent at first.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
This problem is like my avatar. See how it is cyclic and goes offshing point? Very Zen!
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
that is off to a vanishing point.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
You're a very good student, I can tell!
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Thanks! The function part is very interesting, I didn't know it was okay to think of the restrictions like that!
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Math is way cool. Intuitive and analytical at the same time!
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
That's true. I'm always thinking of how math is all about creativity.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
That's it! There is an absolutely great biography about John Nash and his math exploits. It was made into a good movie called "A Beautiful Mind". The written account is way better though. All about the creative thinking process.
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
I'll definitely look it up! Thanks!
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
You'll love it I guarantee it!
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Watch the movie first if you don't have a lot of time, though. Actually a great movie.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
So gotta go, but wonderful working with you and thx for the recognition medal!
 2 years ago

MelindaR Group TitleBest ResponseYou've already chosen the best response.0
Goodbye! Thanks and keep in touch!
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Will do!
 2 years ago
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