Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

appleibeats · Jan 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I get it now.

Except for how you went from 2 + | N - 2 | = 2 + 2 - N

InteGrand · Jan 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I get it now.

Except for how you went from 2 + | N - 2 | = 2 + 2 - N

$\noindent Paradoxica said $N$ was negative. This implies that $N-2$ is also negative, so $|N-2| = -(N-2) = 2-N$.$

appleibeats · Jan 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Use a right angle triangle to show that :

sin(cos^-1 x) = ( 1 - x^2)^1/2

InteGrand · Jan 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Use a right angle triangle to show that :

sin(cos^-1 x) = ( 1 - x^2)^1/2

$First we prove this for $x>0$ (i.e. $0<x<1$, so $\cos ^{-1}x$ is an acute angle (between $0$ and $\frac{\pi}{2}$)). (The case for $x=0$ is easily seen by substitution: $\cos^{-1} 0 = \frac{\pi}{2}$, so the sine of this is $1$, which is indeed equal to $\sqrt{1-0^2}$).$

$Let $\theta = \cos^{-1} x$, $0<x<1$. Then $\theta$ is the angle of a right-angled triangle. We need to show that $\sin \theta = \sqrt{1-x^2}$. Let $\theta$ be $\angle ABC$ in a triangle $ABC$ with right angle at $C$. Since $\theta = \cos^{-1}x$, we have $x=\cos \theta$. So let the adjacent side to $\theta$ have length $x$, then the hypotenuse has length $1$, as $\cos \theta=x$.$

$Then by Pythagoras' theorem, the third side has length $\sqrt{1-x^2}$. Then $\sin \theta = \frac{ \text{opposite} }{ \text{hypotenuse} } = \frac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}$, as required.$

$For the case of $x<0$, we see let $x=-y$, where $y>0$, then $\cos^{-1} x = \pi - \cos^{-1}y$, using the identity $\cos ^{-1} \eta + \cos^{-1} (-\eta) = \pi$. So$

$\begin{align*} \sin \left(\cos^{-1} x\right) &= \sin \left( \pi - \cos^{-1}y \right)\\ &= \sin \left( \cos^{-1}y \right) \quad (\text{using the identity }\sin \left(\pi - \eta \right) = \sin \eta) \\ &= \sqrt{1-y^2} \quad (\text{using what we proved for positive inputs}) \\ &=\sqrt{1-(-x)^2}\\ &=\sqrt{1-x^2}. \end{align*}$

$\noindent This completes the proof.$

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-01-09 at 10.30.18 am.png

How do you do question a)

The answer is 56/65

integral95 · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32773

How do you do question a)

The answer is 56/65

$sin( \alpha + \beta) = sin \alpha cos \beta + sin \beta cos \alpha \\ \\ \therefore sin (sin^{-1} \frac{4}{5} + sin^{-1} \frac{12}{13}) = \\ \\ \\ sin (sin^{-1} \frac{4}{5} )cos(sin^{-1} \frac{12}{13}) + sin(sin^{-1}\frac{12}{13}) cos(sin^{-1} \frac{4}{5}) = \\ \\ \\ (\frac{4}{5}) (\frac{5}{13}) + (\frac {12}{13}) (\frac{3}{5})$

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-01-09 at 11.47.29 am.png

How do you do this question.

I used compound angle formula to get

cos a cos b = sin a sinb

Then plugged in a = inverse sin x

and b = inverse tan x

Not sure where to go from here. Were those the correct initial steps??

InteGrand · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32774

How do you do this question.

I used compound angle formula to get

cos a cos b = sin a sinb

Then plugged in a = inverse sin x

and b = inverse tan x

Not sure where to go from here. Were those the correct initial steps??

$Draw right-angled triangles to get expressions for $\cos \left(\sin^{-1} x\right)$ etc.$

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I have done that, and got

sin a = x

where - pi/2 < x < pi/2

I presume in 1st quad

And drew a diagram of it

now if i do cos a : how do I know what the sides lengths are, I am only given 'x'.

InteGrand · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I have done that, and got

sin a = x

where - pi/2 < x < pi/2

I presume in 1st quad

And drew a diagram of it

now if i do cos a : how do I know what the sides lengths are, I am only given 'x'.

For side lengths, you can let one of them be unit length (length 1) and another x, choosing them so that the sine of the correct angle is x in your diagram. The remaining side can be found via Pythagoras.

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thanks it worked finally.

But now I am having trouble with part b)

So using the equation cos( a + b ) = ..... from a)

I used the fact given that a + b = pi/2 from question

so subing it in and RHS becomes 0

Then I got rid of the fraction on the LHS by multiplying the equation by it

Now I am unsure where to go from here....

Is what I have done correct??

InteGrand · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Thanks it worked finally.

But now I am having trouble with part b)

So using the equation cos( a + b ) = ..... from a)

I used the fact given that a + b = pi/2 from question

so subing it in and RHS becomes 0

Then I got rid of the fraction on the LHS by multiplying the equation by it

Now I am unsure where to go from here....

Is what I have done correct??

$\noindent Well since the fraction is equal to 0, its numerator is 0 (note that its denominator is always positive, so we don't need to worry about potential 0 on the denominator). So we have $\sqrt{1-x^2}-x^2=0$ (since the numerator is $\sqrt{1-x^2}-x^2$). So we have$

$\begin{align*}\sqrt{1-x^2}&= x^2 \\ \Rightarrow \sqrt{1-u}&= u \quad (\text{where we have let }u=x^2) \\ \Rightarrow u^2 &= 1-u \\ \Rightarrow u^2 +u -1 &= 0 \\ \Rightarrow u&= \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2\cdot 1} \quad (\text{quadratic formula}) \\ \Rightarrow u &= \frac{-1\pm \sqrt{5}}{2}.\end{align*}$

$\noindent But $u=x^2$, so $u$ cannot be negative, so we take the positive solution. So $u=x^2 = \frac{-1+\sqrt{5}}{2}$, as required.$

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-01-09 at 3.59.36 pm.png

Could someone explain this function.

Paradoxica · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Useless if you don't know what mod means.

appleibeats · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I don't know what mod is

InteGrand · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32775

Could someone explain this function.

$That function takes in an angle $x$ and returns the `equivalent angle' in the range $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, where by `equivalent angle', I mean the angle $\theta$ whose tan is the same as the tan of the angle $x$. Basically for any input angle $x$, check where on the graph of $\tan x$ this is, and move it to the corresponding point on the tan graph's branch that goes from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. This new $x$-value is the function value.$

integral95 · Jan 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32775

Could someone explain this function.

it's essentially a straight line y = x defined between -pi/2 and pi/2 (not inclusive since that's undefined, so there would be open circles there) followed by the exact same line between pi/2 and 3pi/2, and another between 3pi/2 and 5pi/2 etc.

This is why the period is pi and the range is strictly between -pi/2 and pi/2.

appleibeats · Jan 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-01-10 at 9.25.50 am.png

I can't seem to simplify to get the answer for the second derivative

Maybe my first derivative is wrong?

For the first derivative I got : ( x( tan^-1 x - 1) ) / (1 + x^2)

Is that correct?

InteGrand · Jan 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 32778

I can't seem to simplify to get the answer for the second derivative

Maybe my first derivative is wrong?

For the first derivative I got : ( x( tan^-1 x - 1) ) / (1 + x^2)

Is that correct?

$\noindent Your first derivative should be something like $\tan^{-1} x$, because the second derivative needs to be $\frac{1}{1+x^2}$, and this is the derivative of $\tan^{-1} {x}$.$

appleibeats · Jan 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Find the derivative of in the simplest form :

sin^-1 1/4 . ( 2x - 3)

Answer says 2/ (7 + 12x - 4x^2)^1/2

Not sure what to do with the 1/4
Do you multiply it with the numbers in the brackets??

Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

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