A level Maths 2019 further trig question Watch

Yellowlemon123
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Hi all,
So I'm struggling to work out the answer to part B, C and D. I've attached the picture of the question and my working for the first part of the question.

TIA!
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RDKGames
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(Original post by Yellowlemon123)
Hi all,
So I'm struggling to work out the answer to part B, C and D. I've attached the picture of the question and my working for the first part of the question.

TIA!
For part b (i) you are on the right track...

1 = \alpha - 10\cos(0) + 3\sin(0)

What are the value of cos(0) and sin(0) ?? Hence alpha is...?

For b (ii) you need to incorporate the fact that 10\cos \theta -3\sin \theta = \sqrt{109}\cos(\theta + 16.7) into H = \alpha - (10\cos(80t) - 3\sin(80t))... how can you rewrite H using this??
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simon0
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Also to point out, for your answer to question 1a, the identity is:

 \displaystyle R \cos(a+b) = R \cos(a) \cos(b) - R \sin(a) \boldsymbol{\sin} (b).

However, the rest of your answer for question 1a is fine.
-----------------------------------------------------------------------------------------------------------------------------------------------------------

For question 1c, you found the equation for height  H has form:

 \displaystyle \alpha - \sqrt{109} \cos (t + 16.7^{\circ}),

where  \alpha is to be found.

Can you find the time that the passenger has the greatest height for the first time (call this  t_{\mathrm{max}} ).
(Ideally found by knowing how cos behaves).

Then cos is a periodical function so cos has a maximum/minimum again after  T units (e.g.  \cos(0 + T) = \cos(0) = 1, for a particular value of T).

Are you able to find T and use this to find the time needed (where the passenger is at the maximum height for the second time)?

-----------------------------------------------------------------------------------------------------------------------------------------

For question 1d, if the ferris wheel is faster, you expect more cycles.

How do we get more cycles in a cos graph (under a certain time)?

Does this help?
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Yellowlemon123
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(Original post by RDKGames)
For part b (i) you are on the right track...

1 = \alpha - 10\cos(0) + 3\sin(0)

What are the value of cos(0) and sin(0) ?? Hence alpha is...?

For b (ii) you need to incorporate the fact that 10\cos \theta -3\sin \theta = \sqrt{109}\cos(\theta + 16.7) into H = \alpha - (10\cos(80t) - 3\sin(80t))... how can you rewrite H using this??
So would I have to cos^-1 and sin^-1 (which are 90 and 0) to get alpha or would I use the actual values of cos(0) and sin(0) (which are 1 and 0)? And then for b(ii) I have substituted 80t into sqrt {109}cos (theta + 16.7) so it's sqrt {109}cos(80t + 16.7) is that right?
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RDKGames
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(Original post by Yellowlemon123)
So would I have to cos^-1 and sin^-1 (which are 90 and 0) to get alpha or would I use the actual values of cos(0) and sin(0) (which are 1 and 0)? And then for b(ii) I have substituted 80t into sqrt {109}cos (theta + 16.7) so it's sqrt {109}cos(80t + 16.7) is that right?
Why would you inverse cos or sin?? cos(0)=1 and sin(0)=1 so just stick them in.

Yes its right, so rewrite H in terms of this sqrt {109}cos(80t + 16.7)
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Yellowlemon123
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(Original post by RDKGames)
Why would you inverse cos or sin?? cos(0)=1 and sin(0)=1 so just stick them in.

Yes its right, so rewrite H in terms of this sqrt {109}cos(80t + 16.7)
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Is this right then?
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RDKGames
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(Original post by Yellowlemon123)
Is this right then?
You got the right eqn for H now, but i dont understand what you are doing afterwards.

Since H is of the form
(constant) - (other costant)*(cos function)
you need to realise that max height is given when this result is maximised. Clearly, the two constant stay the same so its just the matter of focusing on (cos function) and seeing what value of it will give us max height. What are the two extreme values of the (cos function) ??
For which do we get max height ??
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Yellowlemon123
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(Original post by RDKGames)
You got the right eqn for H now, but i dont understand what you are doing afterwards.

Since H is of the form
(constant) - (other costant)*(cos function)
you need to realise that max height is given when this result is maximised. Clearly, the two constant stay the same so its just the matter of focusing on (cos function) and seeing what value of it will give us max height. What are the two extreme values of the (cos function) ??
For which of these two do we get max height ??
Would it be + or - 16.7/80 then?
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RDKGames
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(Original post by Yellowlemon123)
Would it be + or - 16.7/80 then?
No... And i precisely ignored the argument because you dont need it.

For y = cos(x) you have focused on x values to give me this answer, but im asking what are the two extreme y values a cos function can take.
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Yellowlemon123
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(Original post by RDKGames)
No... And i precisely ignored the argument because you dont need it.

For y = cos(x) you have focused on x values to give me this answer, but im asking what are the two extreme y values a cos function can take.
Ohhh so it'll be between 1 and - 1
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RDKGames
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(Original post by Yellowlemon123)
Ohhh so it'll be between 1 and - 1
Thats the range but what are the two extreme values of it??

Well theyre just 1 and -1.
So check for which one we get the highest height.
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Yellowlemon123
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(Original post by simon0)
Also to point out, for your answer to question 1a, the identity is:

 \displaystyle R \cos(a+b) = R \cos(a) \cos(b) - R \sin(a) \boldsymbol{\sin} (b).

However, the rest of your answer for question 1a is fine.
-----------------------------------------------------------------------------------------------------------------------------------------------------------

For question 1c, you found the equation for height  H has form:

 \displaystyle \alpha - \sqrt{109} \cos (t + 16.7^{\circ}),

where  \alpha is to be found.

Can you find the time that the passenger has the greatest height for the first time (call this  t_{\mathrm{max}} ).
(Ideally found by knowing how cos behaves).

Then cos is a periodical function so cos has a maximum/minimum again after  T units (e.g.  \cos(0 + T) = \cos(0) = 1, for a particular value of T).

Are you able to find T and use this to find the time needed (where the passenger is at the maximum height for the second time)?

-----------------------------------------------------------------------------------------------------------------------------------------

For question 1d, if the ferris wheel is faster, you expect more cycles.

How do we get more cycles in a cos graph (under a certain time)?

Does this help?
It kind of helps, but still a bit confused about the last 2 questions. (thank you for letting me know about the mistake I made in part a, I didn't realise I'd put cos instead of sin)
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simon0
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To carry on from RDKGames and to finish off question 1bii:

You found the equation for the height of the passenger as:

 H = 11 - \sqrt{109} \cos ( 80t + 16.7) \quad .

(Note this is just a cos graph which is upside down, shifted up by 11 units, stretched by  \sqrt{109} units and has a increased frequency).

Now, you are correct that the cos graph has extremes of 1 and -1.
So H at the minimum/maximum is:

 H = 11 - \sqrt{109} (1), and  H = 11 - \sqrt{109} (-1) .

Now, which one is the maximum and what value is it?
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simon0
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For question 1c, note the graph of:  y = - \cos(t), (where t is in degrees).

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The graph above has a maximum at  t = 180 .

Now I said cos is a periodic function, so what value is x for the next maximum (let us call it "A").

Equation for H is (again):

 11  -\sqrt{109} \cos (80t + 16.7) .

We want the second maximum so do we want the argument in the cos function (of H) to equal to?

So:

 80t + 16.7 = ??? .
Spoiler:
Show
 80t + 16.7 = 180 + ``\text{Some value}'' .

Can you fill in the rest now?
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simon0
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For question 1d:

If the ferris wheel went faster, how will that show on the graph?
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