This is like any other educational corner.
Only, this one will contain lots and lots of numbers compared to the rest
Well, my dear brothers and sisters, do TRY to help each other as we are ordained to do so.
I also hope that those who do not enjoy mathematics will soon come to love it.Its a very challenging subject but it is indeed great fun!!!!:coolsis:
‘Say: If the ocean were ink wherewith to write out the words of my Lord, sooner would the ocean be exhausted, even if We added another ocean like it.’~Al Qu'raan (18:109)
I believe they start calculus your third or fourth(the last) year in high school. So I'd say mainly 16/17/18 ish. On that note, I'm surprised you started at 14/15!
It really depends though which math track you're on. Some people are quite blessed, they go on a faster track. Others don't touch a real calculus until they've graduate their fourth year in high school and gone to a university.
You're saying the final years of high school right? Years 10/11 are our final years, though I covered calculus in more detail when I studied for A-level Maths.
And promptly forgot everything.
At the stationary point P, x = k, and dy/dx = 0, so set the above expression ^^ to equal zero, and replace x with k:
2tan2k + 4xsec^2(2k) - 2sec^2(2k) = 0
Multiply throughout by cos^2(2k) - or divide by sec^2(2k) - to get:
Difficult stuff indeed. But maybe that's because I'm mesmerised by my art to focus on the differentiation. I think I'll take a break and then come back.
Difficult stuff indeed. But maybe that's because I'm mesmerised by my art to focus on the differentiation. I think I'll take a break and then come back.
Lol, yup, good idea...using the product rule gets wayy complicated in this example...
format_quote Originally Posted by Whatsthepoint
x satisfies this equation
cos(2x) + cos(3x) = 1
Show that it also satisfies this one:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)
hEr's one for ya, Raynn
Lol, it looked like this was gonna take foreverr, but it wasn't too bad...so, we're trying to show:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)
Starting with the right side, and used the double angle formula for sinx,
sin4x + 2sin5x + sin6x
= 2sin2xcos2x + 2(sin2xcos3x + sin3xcos2x) + 2sin3xcos3x ......[the only odd angle is the sin5x; divide that into 2x and 3x]
= (2sin2xcos2x + 2sin2xcos3x) + (2sin3xcos2x + 2sin3xcos3x).....[regrouping]
= 2sin2x(cos2x + cos3x) + 2sin3x(cos2x + cos3x)......[factorizing]
We know from the previous identity thing that (cos2x + cos3x) = 1
Therefore,
sin4x + 2sin5x + sin6x = 2sin2x(1) + 2sin3x{1) sin4x + 2sin5x + sin6x = 2sin2x + 2sin3x
Lol, yup, good idea...using the product rule gets wayy complicated in this example...
Lol, it looked like this was gonna take foreverr, but it wasn't too bad...so, we're trying to show:
2sin(2x) + 2sin(3x) = sin(4x) + 2sin(5x) + sin(6x)
Starting with the right side, and used the double angle formula for sinx,
sin4x + 2sin5x + sin6x
= 2sin2xcos2x + 2(sin2xcos3x + sin3xcos2x) + 2sin3xcos3x ......[the only odd angle is the sin5x; divide that into 2x and 3x]
= (2sin2xcos2x + 2sin2xcos3x) + (2sin3xcos2x + 2sin3xcos3x).....[regrouping]
= 2sin2x(cos2x + cos3x) + 2sin3x(cos2x + cos3x)......[factorizing]
We know from the previous identity thing that (cos2x + cos3x) = 1
Therefore,
sin4x + 2sin5x + sin6x = 2sin2x(1) + 2sin3x{1) sin4x + 2sin5x + sin6x = 2sin2x + 2sin3x
There we are!
Yep.
when I was supposed to know it I didn't know all these formulas, like sin (alpha + beta).
You're officially an aspie, Raynn.
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