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AuthorTopic: Maths Q (Read 623 times)
fireydeity
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Maths Q
« Thread Started on May 22, 2006, 12:53pm »
i was wondering how we were supposed to do one of the questions in our maths exam:
cosx(sinx-3cosx)=0
solve 0<x<2Pi in radians to three decimal places
x=theta
can you help?
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Re: Maths Q
« Reply #1 on May 23, 2006, 5:46pm »
When I revise C2 (starting from tmrw after my Stats exam), I'll give it a proper go...
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Re: Maths Q
« Reply #2 on May 28, 2006, 11:43am »
Well i've had a go, i may well be wrong but this is what i think you do=

If you multiply any two numbers together and gat the arwnser zero then one or both of the numbers multiplied together must also equal zero.

Therefor if cosX(sinX-3cosX)=0 then
cosX=0
and/or
sinX-3cosX=0

cosX=o at the points 1.571 and 4.712 to 3 d.p.

if sinX-3cosX=o then sinX=3cosX
this occurs at points 1.249 and 4.391 to 3 d.p.

i havent propely checked it thoug so this may all be wrong,
« Last Edit: May 28, 2006, 12:04pm by Comrade Wiggeh »Link to Post - Back to Top  IP: Logged
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Re: Maths Q
« Reply #3 on May 28, 2006, 9:46pm »
cosx(sinx-3cosx)=0
solve 0<x<2Pi in radians to three decimal places
x=theta

Right... um, well I think maybe...

cosx(sinx-3cosx) = 0
So cosx = 0 or sinx-3cosx = 0

cosx = 0
x = cos-10
x = 1.571c and 4.712c
(in degrees: 90o and 270o)

sinx-3cosx = 0
sinx = 3cosx
sinx / 3cosx = 3cosx / 3cosx
sinx / 3cosx = 1
[using rule that sinx/cosx = tanx]
1/3 tanx = 1
tanx = 3
x = tan-1 3
x = 1.249c and 4.391c
(in degrees: 71.6o and 251.6o)



So I would agree with Wiggeh's answers and I've given my method above.

Other useful trigonemetry stuff, where @ = theta:

sin2@ + cos2@ = 1

sinx / cosx = tanx

Cosine rule [where 'a' is the side of a triangle opposite to angle 'A' and the other sides are 'b' and 'c']:
a2 = b2 + c2 - (2bc x cosA)

Sine rule [using the same labelling as above]:

sin A = sin B = sin C
.. a ...... b ....... c

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Re: Maths Q
« Reply #4 on May 30, 2006, 2:53pm »
not that i have a better answer but i dont think that can be right, if u have to solve between 0 and 2Pi that only gives one value. and surely x would equal the same thing whichever way you did it. i feel it must be somthing to do with the identity sin2x - cos2x=1 but i couldnt get it to work.
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Re: Maths Q
« Reply #5 on Jun 1, 2006, 3:01pm »
i tryed using sin2x - cos2x=1 but it doesnt realy work, @ (x) will equal many things, it is Y that is the same all the way through at 0, if you put the equation into your graphic calculator, in radian mode, you can see it cuts the X axis alot of times, therefor there would be alot of @values, like sin^1(1/2) can be 30 or 150, or at least thats my line of thought.

When i ploted the graph on my graph calculator and found the intersects they where as Barno and i predicted, with a few extra however when i put the others into the equation it came out with syntax error, haven't checked why yet. I could well still be wrong but hay.....
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Re: Maths Q
« Reply #6 on Jun 4, 2006, 6:53pm »

Quote:
not that i have a better answer but i dont think that can be right, if u have to solve between 0 and 2Pi that only gives one value. and surely x would equal the same thing whichever way you did it. i feel it must be somthing to do with the identity sin2x - cos2x=1 but i couldnt get it to work.


Using: sin2x + cos2x = 1

cosx(sinx-3cosx)=0
(cosx*sinx)-3cos2x
cosx*sinx = 3cos2x
cosx*sinx = 3(1 - sin2x)
cosx*sinx = 3 - 3sin2x

Lol... I need to eat pancakes now... I'll get back to this - probably after my exams lol... the suspense is ---- TOOOO GREAT!
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Re: Maths Q
« Reply #8 on Sept 5, 2009, 6:10am »
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