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I can answer qs on maths or physics

Sup dudes and dudettes, I'm free for the next couple of hours so if you are stuck on any maths or physics questions/topic then hmu. I am currently doing physics at Cambridge and will try my best to help :smile:
Reply 1
Original post by Darth_Narwhale
Sup dudes and dudettes, I'm free for the next couple of hours so if you are stuck on any maths or physics questions/topic then hmu. I am currently doing physics at Cambridge and will try my best to help :smile:


If you want to help, just go around and answering unanswered threads. No-one will bother with a thread like this.
Original post by Darth_Narwhale
Sup dudes and dudettes, I'm free for the next couple of hours so if you are stuck on any maths or physics questions/topic then hmu. I am currently doing physics at Cambridge and will try my best to help :smile:


Please could you help me with this trig question?

solve for 0 < x < 360 degrees

2sin(x) = cos(x+60)

This is what i've got so far....

cos(x+60) = cosx cos60 - sinx sin60

i just can't see where i have to go with it next....


Thanks :-)
Original post by ellie12344
Please could you help me with this trig question?

solve for 0 < x < 360 degrees

2sin(x) = cos(x+60)

This is what i've got so far....

cos(x+60) = cosx cos60 - sinx sin60

i just can't see where i have to go with it next....


Thanks :-)


You know what cos (60) and sin(60) are, so you are left with an expression of sin(x) and cos(x). Get all the sin(x) terms on one side of the = and the cos(x) terms on the other side. Then divide by cos(x) to get something along the lines of tan(x) = ...

Then just type it into the calculator. I got 9.9 and 189.9 but I may have made a mistake somewhere :smile:
Reply 4
Unparseable latex formula:

[br] [br]\int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^2}r\,dr\,d\theta = \pi \[br] [br]

Original post by Darth_Narwhale
You know what cos (60) and sin(60) are, so you are left with an expression of sin(x) and cos(x). Get all the sin(x) terms on one side of the = and the cos(x) terms on the other side. Then divide by cos(x) to get something along the lines of tan(x) = ...

Then just type it into the calculator. I got 9.9 and 189.9 but I may have made a mistake somewhere :smile:




Thank you very much!!!!
Original post by Pikachū
[br][br]02π ⁣ ⁣0er2rdrdθ=π [br][br] [br] [br]\int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^2}r\,dr\,d\theta = \pi \ [br] [br]



What the hell is that O.o The only sign I know is the Integral, Change in (d) and Pi.

Oh and the number 0.

What does that formula do.....
(edited 7 years ago)
Original post by Pikachū
[br][br]02π ⁣ ⁣0er2rdrdθ=π [br][br] [br] [br]\int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^2}r\,dr\,d\theta = \pi \ [br] [br]


What on earth is the upper bound for the inner integrand?
Reply 8
Original post by Bulletzone
What the hell is that O.o The only sign I know is the Integral, Change in (d) and Pi.

Oh and the number 0.

What does that formula do.....


Grade 11 Japanese maths :yes:
Original post by Pikachū
Grade 11 Japanese maths :yes:


The hardest Year 11 maths i did was Quadratics :rofl:
Original post by Pikachū
Grade 11 Japanese maths :yes:


That's a gaussian integral in polar, but with an extra factor of r right? If the extra r wasn't there it would just be (pi)^1/2 I think...

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