Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

rand_althor · Oct 7, 2015

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

If you haven't learn the auxiliary method for combining two sine and cosine functions, you can use addition of ordinates. For each x-coordinate, add the y-coordinates of y=sinx and y=cosx. So at x=0, sin(0)=0 and cos(0)=1 -> 1+0=1.
Period: The curve starts at 1, and takes a total of 2pi units before it reaches 1 again (excluding x=pi/2). So the period is 2pi.
Amplitude: On the graph, you can see that the highest point of y=sinx+cosx is when y=sinx and y=cosx are equal. So work out the values of y=sinx and y=cosx when sinx=cosx, and add them to get the highest point. This is your amplitude.

InteGrand · Oct 7, 2015

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

appleibeats said:
I have sketched y = sinx and y = cosx, for 0 </= x </= 2pi

But am unsure how to sketch y = sinx + cosx for 0 </= x </= 2pi.

Then how do you calculate the period and amplitude of y = sinx + cosx

$The period is $2\pi$ and the amplitude is $\sqrt{2}$.$ .

appleibeats · Oct 7, 2015

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

Zlatman said:
Are you using radians for sin x?

EDIT: Once y > 200, x and sin x will have no more solutions. Therefore, we're looking at the domain of x < 200 (since y = x, and y must be less than 200). In this domain, sin x completes 31.8 cycles (200 divided by 2pi), and y = x passes through sin x twice in the first half of each cycle. This gives us 64 solutions. But x = 0 is included in this, and that is not positive; hence we have 63 solutions.

I hope that makes some sense.

Now how do I find a positive integer value of n such that sinx = x/n has 69 positive solutions.

Zlatman · Oct 7, 2015

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

appleibeats said:
Now how do I find a positive integer value of n such that sinx = x/n has 69 positive solutions.

We're dealing with the lines y = x and y = n sin x.

Now, for there to be 69 positive solutions (70 including x = 0), the line y = x must be greater than the line y = n sin x after the 70th solution, which occurs during the first half of the 35th cycle of sin x. The earliest y = x can exceed y = n sin x forever is just after the peak of the first half of the 35th cycle, e.g. after 34.25 cycles. The latest y = x can exceed y = n sin x forever is just before the peak of the first half of the 36th cycle, e.g. after 35.25 cycles.

Multiplying 34.25 and 35.25 by 2pi gives us a positive integer domain of 216 to 221, meaning from this domain onwards, y = x must always be greater than y = n sin x.

Since y = x, y must be between 216 and 221 in this domain.

As n is the largest possible value for y = n sin x, n must be between 216 and 221.

Probably didn't explain that well, but it's almost the reverse of the last method.

EDIT: changed the cycles from 34.5 to 35, to 34.25 to 35.25.

InteGrand · Oct 7, 2015

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

appleibeats said:
Now how do I find a positive integer value of n such that sinx = x/n has 69 positive solutions.

$This is equivalent to asking for 70 non-negative solutions, i.e. including the solution $x=0$. Since there are two solutions per period, with both solutions occurring in the first half of each period, we need there to be at least 34.5 periods. Note that 34.5 periods finish when $x = 34.5\times 2\pi = 69\pi = 216.7698931....$. The line $y =\frac{x}{n}$ needs to stay below $y=1$ for all $x$ in this range, and we would like it to exit and go above $y=1$ before the start of the next period, i.e. before $x = 35\times 2\pi = 219.91148....$ (to ensure we don't have more than 69 positive solutions). This means we may take $n = 219$, so that for all $x \in (0,69\pi)$, the line is between 0 and 1, and the line exits this range before $x = 35\times 2\pi$.$

Edit: Done above

appleibeats · Oct 7, 2015

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

If sinx = 3/4 and pi/2 < x < pi , find the exact value of sin2x.

kawaiipotato · Oct 7, 2015

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

$$ For $ \frac{\pi}{2} < x < \pi ,$

$\pi < 2x < 2 \pi$

$-1 < \sin 2x < 0$

$\sin 2x = 2 \sin x \cos x$

$$ Note that in this domain, $ -1 < \cos x< 0$

$$ Since $ \sin x = \frac{3}{4}, $ then $ \cos x = \frac{ - \sqrt{4^2 - 3^2 }}{4} ( $ by drawing up a right angled triangle $ )$

$\cos x = \frac{- \sqrt{7}}{4}$

$\therefore \sin 2x = 2 * (\frac{3}{4}) * (\frac{- \sqrt{7}}{4})$

$= \frac{ - 3 \sqrt{7}}{8}$

InteGrand · Oct 7, 2015

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

appleibeats said:
If sinx = 3/4 and pi/2 < x < pi , find the exact value of sin2x.

$Note that in this range of $x$, $\cos x<0$, so $\cos x =-\sqrt{1-\sin^2 x} = -\sqrt{ 1-\left(\frac{3}{4}\right)^2} =- \frac{\sqrt{7}}{4}$ Now, $\sin 2x =2\sin x \cos x = 2\times \frac{3}{4}\times \left(-\frac{\sqrt{7}}{4}\right) = -\frac{3\sqrt{7}}{8}$.$

appleibeats · Oct 7, 2015

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

IF sin A = 2/3 , pi/2 < A < pi , and tan B = 2/3 , pi< B < 3pi/2 , show that cos(A + B) = 3(5)^1/2 + 4 / 3(13)^1/2

Instead of root 13 in the denominator, I got the exact same answer except for root 5 in the denominator. Not sure if I am wrong or there is a typo?

kawaiipotato · Oct 7, 2015

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

appleibeats said:
+ 4 / 3(13)^1/2

If you're talking about that part it should be root13. Expanding cos(a+b) = cosAcosB - sinAsinB
sinA = 2/3 and sinB = -2/(13)^(1/2) so -sinAsinB = 4/(3(13)^1/2

appleibeats · Oct 7, 2015

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

Find the exact value of sin2x.

InteGrand · Oct 7, 2015

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

appleibeats said:
Find the exact value of sin2x.

Given what?

appleibeats · Oct 7, 2015

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

InteGrand said:
Given what?

Sorry, wrote the question wrong. Meant to be, find the exact value of sin13pi/12

InteGrand · Oct 7, 2015

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

appleibeats said:
Sorry, wrote the question wrong. Meant to be, find the exact value of sin13pi/12

$\sin \left(\frac{13\pi}{12} \right)=\sin \left( \pi + \frac{\pi}{12}\right)$

$=-\sin \left( \frac{\pi}{12}\right)$

$=-\sin \left( \frac{\theta}{2}\right)$, where $\theta = \frac{\pi}{6}$

$$=-\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$ (by square-rooting the trig. identity $\sin^2 t = \frac{1}{2}-\frac{1}{2}\cos 2t$, and noting that $\sin \left( \frac{\theta}{2}\right) >0$ as the argument is acute, so $\sin \left( \frac{\theta}{2}\right) =\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$, not $-\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$)$

$$=-\sqrt{\frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}}$ (as $\cos \theta = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$)$

$=-\sqrt{\frac{2}{4}-\frac{\sqrt{3}}{4}}$

$$=-\frac{1}{2}\sqrt{2-\sqrt{3}}$.$

appleibeats · Oct 7, 2015

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

InteGrand said:
$\sin \left(\frac{13\pi}{12} \right)=\sin \left( \pi + \frac{\pi}{12}\right)$

$=-\sin \left( \frac{\pi}{12}\right)$

$=-\sin \left( \frac{\theta}{2}\right)$, where $\theta = \frac{\pi}{6}$

$$=-\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$ (by square-rooting the trig. identity $\sin^2 t = \frac{1}{2}-\frac{1}{2}\cos 2t$, and noting that $\sin \left( \frac{\theta}{2}\right) >0$ as the argument is acute, so $\sin \left( \frac{\theta}{2}\right) =\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$, not $-\sqrt{\frac{1}{2}-\frac{1}{2}\cos \theta}$)$

$$=-\sqrt{\frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}}$ (as $\cos \theta = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$)$

$=-\sqrt{\frac{2}{4}-\frac{\sqrt{3}}{4}}$

$$=-\frac{1}{2}\sqrt{2-\sqrt{3}}$.$

Unsure about which trig identity you refer to. The answer at the back says 1 - 3^1/2 /2(2)^1/2

InteGrand · Oct 7, 2015

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

appleibeats said:
Unsure about which trig identity you refer to. The answer at the back says 1 - 3^1/2 /2(2)^1/2

$The answer at the back (namely $\frac{1-\sqrt{3}}{2\sqrt{2}}$) can be shown to be equal to the one I gave. One way to do this would be to square my answer, square the book's answer, and observe that they both give the same result, and also observe that both our answers have the same sign, which means they must be equal (because if $x^2=y^2$ and $x$ and $y$ have the same sign, then $x=y$).$

InteGrand · Oct 7, 2015

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

appleibeats said:
Unsure about which trig identity you refer to. The answer at the back says 1 - 3^1/2 /2(2)^1/2

$The trig. identity I was referring to is the half-angle identity for sine: $\sin^2 t = \frac{1}{2}\left(1 - \cos 2t\right)$.$

appleibeats · Oct 7, 2015

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

Find the value of tan 135 degrees

doing tan( 90 + 45 ) gives me the answer 1

but doing tan ( 180 - 45) gives the answer - 1

The correct answer is -1.

Can't find my mistake in the first method. Are both method's valid??

InteGrand · Oct 7, 2015

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

appleibeats said:
Find the value of tan 135 degrees

doing tan( 90 + 45 ) gives me the answer 1

but doing tan ( 180 - 45) gives the answer - 1

The correct answer is -1.

Can't find my mistake in the first method. Are both method's valid??

$The following was \underline{invalid}: $\tan \left(90^\circ + 45^\circ \right) = \tan 45^\circ$. The identity is $\tan \left(180^\circ + t\right) = \tan t$, \emph{not} $\tan \left( 90^\circ + t\right) = \tan t$. The following is valid though: $\tan \left(90^\circ + t\right) = -\tan \left(180^\circ - \left(90^\circ + t\right) \right) = -\tan \left(90^\circ -t \right)=-\cot t$.$

Paradoxica · Oct 7, 2015

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

appleibeats said:
Find the value of tan 135 degrees

doing tan( 90 + 45 ) gives me the answer 1

but doing tan ( 180 - 45) gives the answer - 1

The correct answer is -1.

Can't find my mistake in the first method. Are both method's valid??

tan(90) is undefned.

Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

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