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These two can't be simultaneously correct identities. The first one is the one that is correct. Rearrange that to obtain: sin(2x) = 2*sin(x)cos(x).

^ A quick way to check that that can't be right is to note that that would imply that cos(a)/2 = 1 whenever sin(a) is not 0. But that would require cos(a) = 2, which we know can't happen for real a, since cos(a) is between -1 and 1.

I put up a method above. Is that the method you were using? If not, what did you do?

The rule is sin(A)cos(A) = (1/2)*sin(2A).

For part b), we are given dV/dt in terms of h. We can also find dV/dh in terms of h using geometry. So we can use the chain rule to get dh/dt as a function of h, f(h). So dt will be dh/f(h). So you can find the required time by integrating the RHS from h = 0 to 100.

The rule sin(A)cos(A) = (1/2)sin(2A), which is what you used to get up to what you were up to.

Apply that rule again and it becomes (1/4)*sin(4x).

Re: HSC 2016 2U Marathon No. Only Russell and Whitehead did probably. :p

$\noindent Since $f$ is differentiable at $a$, we have $f^{\prime} (a) = L$ for some real number $L$. We recall (see link below) using symmetric derivatives that $L = \lim _{\delta \to 0} \frac{f(a+\delta) -f(a-\delta)}{2\delta}$. Now, use the substitution $\delta = ph$, $p\neq 0$, $p$ real...

$\noindent For your latest Q., recall that a function can only be differentiable at a point if it is continuous there. Since $f(x)\to 0$ as $x \to 0^{+}$, we also need (and it is sufficient for) the left limit to be 0 for continuity. So we must have $b=0$ (so that the left limit is 0)$.$...

For the MVT one, I think the user laters asked the same Q. It was answered here: http://community.boredofstudies.org/238/extracurricular-topics/349815/mean-value-theorem-proofs.html#post7141393 .

Re: HSC 2016 2U Marathon There's like a 300+ page proof of 1 + 1 = 2.

Re: HSC 2016 3U Marathon It's possible that the question intended to ask what's the distance travelled from one end to the other? Anyway, as asked (distance in a full cycle), the answer has to be 4A, where A is the amplitude.

Is the question asking to sketch the graph of y = 2|x|? To do this, the graph is that of y = 2x for x >= 0, and y = -2x (downward sloping line) for x < 0. The graph is sketched below: http://wolframalpha.com/input/?i=Sketch+the+graph+of+y+%3D+2%7Cx%7C&x=0&y=0

Yes it works: sin^2 x = (sin(x))^2 = (cos(90 deg - x))^2 = cos^2 (90 deg - x).

= 1 by Pythagorean trig. identity, because sin^2(40 deg) = cos^2(50 deg) (complementary angles identity).

I think the same Q as part c) was asked here and answered: http://community.boredofstudies.org/238/extracurricular-topics/349271/laters-maths-help-thread.html#post7133125.

a) We can use a function whose graph dips below the x-axis and above it and attains max and min. like that, before tending to 0. You can use this idea to find the formula for such a function. b) We can use a function that has a minimum below the x-axis but doesn't reach any max. E.g. f(x) =...

The top half of the question seems to be missing.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread The distance from the origin is greatest at x = -4. The displacement includes the sign, so is greatest at x = +2.