Cambridge Prelim MX1 Textbook Marathon/Q&A (7 Viewers)

InteGrand · Oct 7, 2015

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

Paradoxica said:
tan(90) is undefned.

$That was not the flaw in the logic, the flaw was the assumption that $\tan (90^\circ + 45^\circ)$ was the same as $\tan 45^\circ$, which was probably based on the assumption that the following is a trig. identity: $\tan (90^\circ + \alpha ) = \tan \alpha$ (this is NOT true, however).$

Paradoxica · Oct 7, 2015

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

InteGrand said:
$That was not the flaw in the logic, the flaw was the assumption that $\tan (90^\circ + 45^\circ)$ was the same as $\tan 45^\circ$, which was probably based on the assumption that the following is a trig. identity: $\tan (90^\circ + \alpha ) = \tan \alpha$ (this is NOT true, however).$

oh. I thought he expanded using compound angles.

kawaiipotato · Oct 7, 2015

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

^
Could have also used the idenity
Tan (90+x) = tan (90-(-x))= cot (-x) = -cotx

appleibeats · Oct 8, 2015

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

So the question says Find the acute angle and the obtuse angle between the lines:

2x - y + 1 = 0 and 11x - ( 8 + 5squaroot3)y + 5sqrt3 = 0

So I get Thetha = 30 and that is obviously the acute angle

Now finding the obtuse angle, is it as simple as saying 180 - 30 , so 150

The answer is 30, and 150

or is there more working necessary for obtuse angle find ??

si2136 · Oct 8, 2015

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

appleibeats said:
So the question says Find the acute angle and the obtuse angle between the lines:

2x - y + 1 = 0 and 11x - ( 8 + 5squaroot3)y + 5sqrt3 = 0

So I get Thetha = 30 and that is obviously the acute angle

Now finding the obtuse angle, is it as simple as saying 180 - 30 , so 150

The answer is 30, and 150

or is there more working necessary for obtuse angle find ??

That's it. But did you use the formula to find out the acute/obtuse angle?

appleibeats · Oct 12, 2015

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

what is lim E -> 0 of E^2 / cos^2 E

The answer says 0.

Why is that?? And not 1 ??

InteGrand · Oct 12, 2015

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

appleibeats said:
what is lim E -> 0 of E^2 / cos^2 E

The answer says 0.

Why is that?? And not 1 ??

$Because as $E\to 0$, the numerator goes to 0 and the denominator goes to 1, so the overall fraction goes to $\frac{0}{1}=0$.$

appleibeats · Oct 13, 2015

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

Find dy/dx in terms of x and y by differentiating implicitly:

sin x + cos y = 1

InteGrand · Oct 13, 2015

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

appleibeats said:
Find dy/dx in terms of x and y by differentiating implicitly:

sin x + cos y = 1

$\sin x + \cos y = 1$

$$\Rightarrow \cos x -\sin y \cdot y^\prime = 0$ (differentiating both sides with respect to $x$)$

$$\Rightarrow y^\prime =\frac{\cos x}{\sin y}$.$

kawaiipotato · Oct 13, 2015

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

Differentiating with respect to x means differentiating another variable will give the change in variable with respect to x
So differentiating a y will give dy/dx
sinx + cosy = 1
Diff. Wrt x
d(sinx)/dx + d(cosy)/dx = d(1)/dx
==> cosx - siny * (dy/dx) = 0
dy/dx = cosx/siny

InteGrand · Oct 13, 2015

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

Of course this is just the chain rule:

$$\frac{\mathrm{d} (\cos y)}{\mathrm{d} x} =\frac{\mathrm{d}(\cos y)}{\mathrm{d}y}\times \frac{\mathrm{d}y}{\mathrm{d}x} =-\sin y \cdot \frac{\mathrm{d}y}{\mathrm{d}x}$.$

appleibeats · Oct 13, 2015

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

Given the parametric equations x = e^t cost , y = e^t sint, show that dy/dx = tan ( t + pi/4 ) .

kawaiipotato · Oct 13, 2015

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

Note that y/x = tanx
Aim: show dy/dx = tan (pi/4 + t)
Note the tan expansion: tan (a+b) = (tana + tanb)/(1-tanatanb)
Hence tan (pi/4 + t) = (1 + tant)/(1-tant)
So now your aim is to show:
dy/dx = (1+ y/x)/(1-y/x) = (x+y)/(x-y)

dy/dx = dy/dt * dt/dx
dy/dt = e^t sint + e^t cost (product rule )
dx/dt = e^t cost - e^t sint (product rule)

So dy/dx = (e^t sint + e^t cost)/(e^t cost - e^t sint)
= (x+y)/(x-y) as required

appleibeats · Oct 13, 2015

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

Use the expansion of cos(A+ B) and cos(A-B) to prove that

cosS - cosT = -2sin1/2 (S + T)sin1/2 (S -T)

InteGrand · Oct 13, 2015

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

appleibeats said:
Use the expansion of cos(A+ B) and cos(A-B) to prove that

cosS - cosT = -2sin1/2 (S + T)sin1/2 (S -T)

It's proved here: https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Proof_of_cosine_identities

Scroll up a little bit to see what they substituted θ and ϕ as.

appleibeats · Oct 13, 2015

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

How do you find the derivative of

y = log base e ( cosx)

InteGrand · Oct 13, 2015

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

appleibeats said:
How do you find the derivative of

y = log base e ( cosx)

$In general, $\frac{\mathrm{d}}{\mathrm{d}x}\left( \ln (f(x))\right) = \frac{f^\prime (x)}{f(x)}$ (just from the chain rule).$

$$y=\ln (\cos x)\Rightarrow y^\prime = \frac{\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)}{\cos x}=\frac{-\sin x}{\cos x}=\boxed{-\tan x}$.$

appleibeats · Oct 13, 2015

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

what is the derivative of cos^5 3x .

InteGrand · Oct 13, 2015

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

appleibeats said:
what is the derivative of cos^5 3x .

$Let $y=\cos^5 3x=(\cos 3x)^5$. By the chain rule,$

$\frac{\mathrm{d}y}{\mathrm{d}x}=5(\cos 3x)^4 \times \frac{\mathrm{d}}{\mathrm{d}x}\left( \cos 3x\right)$

$=5(\cos 3x)^4 \times (-\sin 3x)\times 3$

$$=-15\cos^4 (3x) \sin (3x)$.$

appleibeats · Oct 13, 2015

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

So we know x degree is xpi/180 rad

Hence find:

Derivative of (tan (x degrees + 45 degrees))

Cambridge Prelim MX1 Textbook Marathon/Q&A (7 Viewers)

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