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

rand_althor

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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

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

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
.
 

appleibeats

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

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

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

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.
 
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appleibeats

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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

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

















 
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appleibeats

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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

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

+ 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

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

Find the exact value of sin2x.
 

appleibeats

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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

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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??
 

Paradoxica

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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??
tan(90) is undefned.
 

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