The Edexcel C3 (14/06/12 - AM) Revision Thread

$y = \arccos(x)$

Take the cos of both sides

$\cos y = x$

We know $\cos y = \sin \left( \dfrac{\pi}2 - y \right)$

Hence, $\displaystyle \sin \left( \frac{\pi}2 - y \right) = x \implies \frac{\pi}2 - y = \arcsin(x)$

You both are wrong, see my post.

That is so weird ..at 1:25 I edited...at 1:25 you posted this --


em how can we differentiate

$sin^3 x [br]cos^3 x [br]tan^3 x$

May be we both did it at the same time.

For these questions use the chain rule.

$y = [f(x)]^n \\ \dfrac{dy}{dx} = nf'(x)[f(x)]^{n-1}$

I will do one question for you.

$y = \cos^3 x = ( \cos x)^3$

Differentiating it gives, $\dfrac{dy}{dx} = 3 \times ( \cos x)^{3-1} \times (- \sin x ) = - 3 \sin x \cos ^2 x$

Thanks .
Yup I understood that section, but what I was asking was why using sin(y+π/2) on a graph transformation would not give the correct answer. Sorry if I didn't phrase that clearly.

I was meant to ask a totally different question..was gonna say what is the integration of those- might as well ask them later on for C4 page then

We usually write it as $\sin \left( \dfrac{\pi}2 -x \right) = \cos(x) \text{ and } \cos \left( \dfrac{\pi}2 - x \right) = \sin (x)$

I can't think of the answer to your question right now.

The results are usually quoted as 90-x in the books.

Hey people.
Can someone help with Review Exercise 1, Q8 (Edexcel C3 Book) i know how to do be but i don't know how to use the graph to find the function of f. Thanks

help me with this question please!

Use the identities for (sinA + sinB) and (cosA + cosB) to prove that


sin 2x + sin 2y
_______________ = tan (x+y)
cos 2x + cos 2y

SinP + Sin Q = 2* sin[(P+Q)/2 ] * cos[(P-Q)/2 ]

and CosP + Cos Q =2 * cos[(P+Q)/2 ] * cos[(P-Q)/2 ]


P=2x , Q=2y

using this you get


2 sin(x+y) cos(x-y)
--------------------------
2 cos(x+y) cos (x-y)



sin(x+y)
-----------
cos(x+y)


tan(x+y)

That's alright . Thank you!

$\displaystyle y = \arccos (x) \implies \cos (y) = x \implies x = \sin \left( \dfrac{\pi}2 + y \right) \\ \implies \boxed{ \arcsin (x) = \dfrac{\pi}2 + y }$

The range of $y$ is $0\leq y\leq \pi$

So this method gives, $\displaystyle \frac{\pi}{2}\leq \arcsin x \leq \frac{3\pi}{2}$

While the range of $\arcsin (x)$ is $\displaystyle - \frac{\pi}2 \le \arcsin (x) \le \frac{\pi}2$

Hence, we can't do this.

There is no need to draw a graph.

$f(x) = \dfrac{2x+3}{x-1} = \dfrac{2(x-1)+5}{x-1} = \dfrac{2(x-1)}{x-1} + \dfrac5{x-1} = 2 + \dfrac5{x-1}$

We know the domain of $f(x)$ is $x > 1$

So by looking at $f(x) = 2 +\dfrac5{x-1}$ we can deduce that the range of $f(x)$ is $f(x) > 2$

The range of $f(x)$ is the domain of $f'(x)$

Hence the domain of $f'(x)$ is $x > 2$

Hope it makes sense

Hey people. This things been bugging me.
It says in the C3 book, in the example find


f(x) = f-1 (x)

then it solves it for you, but it just says to make it easier solving it goes straight to f(x) = x

how does this work?

None as such but you will need to be able to show all of the double angle formulae and factor formulae from the trig expansions.

Also it is possible you may have to prove some of the standard derrivatives in the formula book by using the derrivatives of sin and cos

Because the range of f(x) is f(x) > 2

It's well worth doing a quick sketch rather than trying to 'plug in' numbers. As stated in in the previous post, I think changing the fraction makes it much more obvious to see where the asymptotes are. So for the inverse function

$f^{-1}(x)=\frac {x+3}{x-2} \equiv 1 + \frac {5}{x-2}$

It's obvious that there's horizontal asymptote at y=1 and a vertical asymptote at x=2.