Maths Homework
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It's been awhile but...
i'd substitute 1-sin^2x for cos^2x and then simplify
u get...
**** i dunno, i've got calc 1 next semester too
Re: Re: Maths Homework
1-sin²X=cos²X ? Is that right? I don't think our maths teacher showed us that...
Originally posted by Chrysaor
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It's been awhile but...
i'd substitute 1-sin^2x for cos^2x and then simplify
u get...
**** i dunno, i've got calc 1 next semester too
1-sin²X=cos²X ? Is that right? I don't think our maths teacher showed us that...
cos^2x + sin^2x = 1
therefore cos^2x = 1 - sin^2x
that's all i got tho too lazy to dig up ol trig and calc notes
they be dusty
therefore cos^2x = 1 - sin^2x
that's all i got tho
too lazy to dig up ol trig and calc notes they be dusty
now prove that the divergence of a gradient field is a self adjoint operator. and that the laplace operator is invariant in any space R(n)
I actually saw no sense in replying to a question already answered, but I did regret that I missed out on a chance to show off my leet math skillz and since this seems to have become the unofficial "Show off your scientific background here 3" thread, I'll just post how I came to answer the question:
When I saw the equaltion I instinctively felt something was wrong with it, and that this function could only be true at singular points. After a moment of thought it was clear:
sin(x)-1 = cos²(x) | substitute a = b for oversight
but: sin(x) <= 1 always => sin(x)-1 <= 0 always
cos²(x) >= 0 always
so we got a <= 0 <= b and want to find a = b. Get it? only a=b=0 is possible, so sin(x)=1 and cos(x)=0
this is true only for (2*y+0.5)*pi, in other words 0.5*pi, 2.5*pi etc.
So 90° is the correct answer.
Now as I mentioned already, the answer was given long ago. This is just showing off my thinking.
Looks more awkward than my actual thougts actually, which were basically "Wait, sin(x) has to be 1 or it'd be negative!" the rest is so deeply indoctrinated I don't percieve it consciously.
When I saw the equaltion I instinctively felt something was wrong with it, and that this function could only be true at singular points. After a moment of thought it was clear:
sin(x)-1 = cos²(x) | substitute a = b for oversight
but: sin(x) <= 1 always => sin(x)-1 <= 0 always
cos²(x) >= 0 always
so we got a <= 0 <= b and want to find a = b. Get it? only a=b=0 is possible, so sin(x)=1 and cos(x)=0
this is true only for (2*y+0.5)*pi, in other words 0.5*pi, 2.5*pi etc.
So 90° is the correct answer.
Now as I mentioned already, the answer was given long ago. This is just showing off my thinking.
Looks more awkward than my actual thougts actually, which were basically "Wait, sin(x) has to be 1 or it'd be negative!" the rest is so deeply indoctrinated I don't percieve it consciously.
OK claw, now you've got your chance to prove yourself
Find all values of x when 0<x<360 for:
9sin²x - 6sin x + cos² x = 0
I think that's the only other one I can't do out of the 35. Then I'm done with trig. identities for good
That is ... until the exams
Find all values of x when 0<x<360 for:
9sin²x - 6sin x + cos² x = 0
I think that's the only other one I can't do out of the 35. Then I'm done with trig. identities for good
That is ... until the exams
Originally posted by Bot_40
OK claw, now you've got your chance to prove yourself
Find all values of x when 0<x<360 for:
9sin²x - 6sin x + cos² x = 0
I think that's the only other one I can't do out of the 35. Then I'm done with trig. identities for good
That is ... until the exams
Hahahah... good luck with that one.
If you would graph the sine and cosine funtions, you'd see the solution (*cough*) was obvious.
Good luck!
Originally posted by Bot_40
I'm done with trig. identities for good
That is ... until the exams
Only if you aren't getting any more math education. I HATE trig. (Because of those damnable identities that I never learned.
) Like anything else you've had before in math, you aren't ever going to be free of it.
Calculus and trig are evil 
I had this **** last year and all I remember is me on a beach with some really sexy girls, I think I was dreaming
so my question is
cos^2 90 = 0
but what is
cos^3 90 = ?
cos^4 90 = ?
so my question is
cos^2 90 = 0
but what is
cos^3 90 = ?
cos^4 90 = ?