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Year 13 Maths Help Thread

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Original post by notnek
That angle should be 22 not 68:


I may have confused you earlier when I mentioned the 68 degree angle, which is actually the other angle in the triangle.


Oh right, why is 22 degrees now inside the triangle?
Original post by osayukiigbinoba
Oh right, why is 22 degrees now inside the triangle?

Look at what I've done in the diagram. I've moved the P force to create a triangle of forces and the angle is 22.
Original post by notnek
Look at what I've done in the diagram. I've moved the P force to create a triangle of forces and the angle is 22.


Ok thank you. How do I start the next question? Sorry I'm struggling with this topic.
Original post by osayukiigbinoba
Ok thank you. How do I start the next question? Sorry I'm struggling with this topic.

For this topic you need to move around the forces to make a triangle of forces:



Can you see what one of the angles in the triangle must be? Then can you use this to find theta?
(edited 7 years ago)
Original post by notnek
For this topic you need to move around the forces to make a triangle of forces:



Can you see what one of the angles in the triangle must be? Then can you use this to find theta?


Is the angle in the bottom right corner 30 degrees?
Original post by osayukiigbinoba
Is the angle in the bottom right corner 30 degrees?

Yes, because they are alternate angles.

You can now use trigonometry for the rest of the question.
Original post by notnek
Yes, because they are alternate angles.

You can now use trigonometry for the rest of the question.


Thank you so much!
Does anyone know how to differentiate (lnx)(lnx)?
Original post by the_malis
Does anyone know how to differentiate (lnx)(lnx)?


You know that d(lnx)/dx is 1/x so apply the product rule.
Or us chain rule with u=lnx and diff u^2
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@Zacken any tips for fp2 polar stuff? Feels weird after having done everything in Cartesian in all my time doing maths
Original post by metrize
@Zacken any tips for fp2 polar stuff? Feels weird after having done everything in Cartesian in all my time doing maths


You'll get used to it after some practice. Surely you worked with polars in complex numbers in whatever module you learnt them? Was it FP1, I don't remember.
Original post by Zacken
You'll get used to it after some practice. Surely you worked with polars in complex numbers in whatever module you learnt them? Was it FP1, I don't remember.


Oh yeah, so I have. Thanks, I'll keep working it at it
Never mind my bad
(edited 7 years ago)
I'm stuck at a differentiation q:

3)a) The point P lies on a curve with the equation y + 1/2 = sin(x + y) + pi/3
P is the point (k, pi/3). Find the smallest possible positive value of k.

This is what I've done so far (I don't even know if it's correct or not):
I'm at a dead end right now
IMG_1195.jpg
Original post by Namita Gurung
I'm stuck at a differentiation q:

3)a) The point P lies on a curve with the equation y + 1/2 = sin(x + y) + pi/3
P is the point (k, pi/3). Find the smallest possible positive value of k.

This is what I've done so far (I don't even know if it's correct or not):
I'm at a dead end right now
IMG_1195.jpg


Do you think you're too prestige to make a thread and ask for help??? smh....

Also, in your workings you have y+12=sin(xy)+π3 y + \frac{1}{2} = \sin(x-y) + \frac{\pi}{3} but the question you have given to us is y+12=sin(x+y)+π3y + \frac{1}{2} = \sin(x + y) + \frac{\pi}{3}

Do you know how to express trig functions of the form acosx+bsin a\cos x + b\sin in harmonic form?
Original post by Naruke
Do you think you're too prestige to make a thread and ask for help??? smh....

Also, in your workings you have y+12=sin(xy)+π3 y + \frac{1}{2} = \sin(x-y) + \frac{\pi}{3} but the question you have given to us is y+12=sin(x+y)+π3y + \frac{1}{2} = \sin(x + y) + \frac{\pi}{3}

Do you know how to express trig functions of the form acosx+bsin a\cos x + b\sin in harmonic form?


Oh soz it was a typo i meant what I've got in my workings
Harmonic form? I'm not sure?
Original post by Namita Gurung
Oh soz it was a typo i meant what I've got in my workings
Harmonic form? I'm not sure?


Unparseable latex formula:

R\cos(\theta \pm \alpha) & R\sin(\theta \pm \beta)

??
Original post by Naruke
Unparseable latex formula:

R\cos(\theta \pm \alpha) & R\sin(\theta \pm \beta)

??


Oh that form. Won't I need another equation for that?
I tried to redo the q and I ended up with something strange. Could someone look through my method?

image.jpg
Original post by Naruke
Do you know how to express trig functions of the form acosx+bsin a\cos x + b\sin in harmonic form?


Original post by Namita Gurung
Oh soz it was a typo i meant what I've got in my workings
Harmonic form? I'm not sure?


Okay guys, calm down. You're both overly complicating this.

You have π3+12=sin(kπ3)+π3\displaystyle \frac{\pi}{3} + \frac{1}{2} = \sin \left(k - \frac{\pi}{3}\right) + \frac{\pi}{3}, subtract π/3\pi/3 from both sides: sin(kπ3)=12\displaystyle \sin \left(k - \frac{\pi}{3}\right) = \frac{1}{2}

If you still can't see what to do, set u=kπ3u = k - \frac{\pi}{3} then you're trying to solve sinu=12\sin u = \frac{1}{2}, this doesn't need any expanding or harmonic form or whatever stuff.

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