tegwin
|
| posted on 7/2/11 at 03:22 PM |
|
|
OT... A maths puzzler
Anyone any good with trigonometric formulae?!
I have two functions:
f(t)=10cos(16t+pi)
g(t)=10cos(20t-pi)
I need to add them together such that h(t)=f(t)+g(t)
Once they are added together I can then differentiate them.... but I am struggling to add them together easily...
I know that cosA+cosB= -2sin(A+B/2) sin(A-B/2) but that does not seem to help...
Anyone any ideas atall?!?... its doing my turnip in!
[Edited on 7/2/11 by tegwin]
------------------------------------------------------------------------------------------------------------------------
Would the last person who leaves the country please switch off the lights and close the door!
www.verticalhorizonsmedia.tv
|
|
|
|
|
blakep82
|
| posted on 7/2/11 at 03:27 PM |
|
|
i have no idea if this works lol
http://www.wolframalpha.com/input/?i=h%28t%29%3D+10cos%2816t%2Bpi%29+%2B+10cos%2820t-pi%29+
________________________
IVA manual link http://www.businesslink.gov.uk/bdotg/action/detail?type=RESOURCES&itemId=1081997083
don't write OT on a new thread title, you're creating the topic, everything you write is very much ON topic!
|
|
|
beagley
|
| posted on 7/2/11 at 03:38 PM |
|
|
Well......
You can factor out the 10 from both f(t) and g(t) which would give you the following.....
A = 16t + pi
B = 20t + pi
10 (cos A + cos B) = -2 * sin (A+(B/2)) * sin (A - (B/2))
which could then be reduced to
cos A + cos B = (-1/5) * (sin(A+(B/2)) * sin(A-(B/2)))
You should also be able to do another substitution on the (A+(B/2)) and the (A-(B/2)) parts of the "sin" functions. So it would read
(-1/5) * (sin X) * (sin Y) where
X = A + (B/2) from above
Y = A - (B/2) from above
Does that give you an equasion that you can differentiate?
I'm not scared!!! I'm just marking my territory.
|
|
|
BenTyreman
|
| posted on 7/2/11 at 03:55 PM |
|
|
No reason why you can't differentiate without combining the trig functions.
f(t)=10cos(16t+pi)
g(t)=10cos(20t-pi)
h(t) = f(t) + g(t) = 10cos(16t+pi) + 10cos(20t-pi)
cos(theta + pi) = cos(theta - pi) = -cos(theta)
so
h(t) = -10cos(16t) - 10cos(20t)
cos'(x) = -x'sin(x)
so
h(t)' = 160sin(16t) + 200sin(20t)
Hope that helps in some way.
|
|
|
simonwinn
|
| posted on 7/2/11 at 06:22 PM |
|
|
My thoughts would be
h(t) = f(t)+g(t) = 10cos(16t+pi) + 10cos(20t-pi)
Factorise the rhs
h(t) = 10cos ((16t+pi) + (20t-pi))
Then simplify the brackets
h(t) = 10cos (16t+20t+pi-pi)
h(t) = 10 cos(36t)
I think this is right, let us know....
|
|
|
Liam
|
| posted on 7/2/11 at 08:47 PM |
|
|
^^ no can do.
cos(a) + cos(b) does not equal cos(a+b) 
+1 for what Ben said - no need to convert them to any other form. Doing that will only make it more complicated to differentiate.
|
|
|
tegwin
|
| posted on 8/2/11 at 09:39 PM |
|
|
hmm... thanks for the input guys, still trying to figure this one out... its a fiddly little bugger...
------------------------------------------------------------------------------------------------------------------------
Would the last person who leaves the country please switch off the lights and close the door!
www.verticalhorizonsmedia.tv
|
|
|
prawnabie
|
| posted on 8/2/11 at 09:52 PM |
|
|
Bugger that, just turn the calculator upside down and spell boobies!
|
|
|
BenTyreman
|
| posted on 8/2/11 at 09:58 PM |
|
|
Which part can't you figure out?
You can differentiate the equation as it stands using the sum rule, the chain rule and the known derivatives of trig functions. The pi can be removed
from the trig functions using simple trig identities.
|
|
|
McLannahan
|
| posted on 8/2/11 at 09:59 PM |
|
|
quote:
Originally posted by prawnabie
Bugger that, just turn the calculator upside down and spell boobies!
Ahh good call PrawnToast.
I'll raise your 5318008 with a 55378008
|
|
|
BenTyreman
|
| posted on 8/2/11 at 10:12 PM |
|
|
If you are desperate to join the two parts, then using the identity cos(X + pi) = cos(X - pi) = -cos(X)
f(t)=10cos(16t+pi) = -10cos(16t)
g(t)=10cos(20t-pi) = -10cos(20t)
h(t) = f(t) + g(t) = -10(cos(16t) + cos(25t))
Now cos(A+/-B) = cos(A)cos(B) -/+ sin(A)sin(B)
Say A = 18t and B = 2t
cos(18t - 2t) = cos(18t)cos(2t) + sin(18t)sin(2t)
and
cos(18t + 2t) = cos(18t)cos(2t) - sin(18t)sin(2t)
so
h(t) = -10(sin(16t) + sin(2t)) = -10(cos(18t)cos(2t) + sin(18t)sin(2t) + cos(18t)cos(2t) - sin(18t)sin(2t))
h(t) = -10(cos(18t)cos(2t) + cos(18t)cos(2t)) = -10cos(18t) * (cos(2t) + cos(2t)) = -20cos(18t)cos(2t)
As far as I am aware, this is as condensed as possible. This can then be differentiated using the product rule and the chain rule.
[Edited on 8/2/11 by BenTyreman]
[Edited on 10/2/11 by BenTyreman]
|
|
|
tegwin
|
| posted on 10/2/11 at 02:34 PM |
|
|
Chears Ben... I think that is actually making some sence.... Its such an illogical topic!
------------------------------------------------------------------------------------------------------------------------
Would the last person who leaves the country please switch off the lights and close the door!
www.verticalhorizonsmedia.tv
|
|
|
Liam
|
| posted on 10/2/11 at 06:23 PM |
|
|
Ben's first answer is by far the easiest way to go about it. Second version above is harder to differentiate.
|
|
|
prawnabie
|
| posted on 10/2/11 at 06:31 PM |
|
|
quote:
Originally posted by McLannahan
quote:
Originally posted by prawnabie
Bugger that, just turn the calculator upside down and spell boobies!
Ahh good call PrawnToast.
I'll raise your 5318008 with a 55378008
Thats me out im afraid!
|
|
|
