# C3 Maths help- Trig

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Hi everyone,

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.

Attachment 736510

Couple of reasons,

1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?

2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi

But in the MS you just do

pi t/5 = 4pi

Help would be appreciated!

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.

Attachment 736510

Couple of reasons,

1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?

2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi

But in the MS you just do

pi t/5 = 4pi

Help would be appreciated!

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#2

(Original post by

Hi everyone,

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.

Attachment 736510

Couple of reasons,

1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?

2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi

But in the MS you just do

pi t/5 = 4pi

Help would be appreciated!

**Mazza2000**)Hi everyone,

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.

Attachment 736510

Couple of reasons,

1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?

2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi

But in the MS you just do

pi t/5 = 4pi

Help would be appreciated!

But for 1), it depends on where 'it' starts - imagine a circle, the time to reach the top (0 degrees)would differ depending on whether you start at 180 or 270.

2) tricky as I can't see the question, but it seems like you did an extra step by using the harmonic identity; the question would be, why? Again, guesswork as I can't see the question, but the difference is between

**time taken**vs

**at what time is the second revolution complete**

(This paper is horrible)

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(Original post by

I remember this Ferris wheel question, but because TSR is silly the attachment isn't working for some reason.

But for 1), it depends on where 'it' starts - imagine a circle, the time to reach the top (0 degrees)would differ depending on whether you start at 180 or 270.

2) tricky as I can't see the question, but it seems like you did an extra step by using the harmonic identity; the question would be, why? Again, guesswork as I can't see the question, but the difference is between

(This paper is horrible)

**Kevin De Bruyne**)I remember this Ferris wheel question, but because TSR is silly the attachment isn't working for some reason.

But for 1), it depends on where 'it' starts - imagine a circle, the time to reach the top (0 degrees)would differ depending on whether you start at 180 or 270.

2) tricky as I can't see the question, but it seems like you did an extra step by using the harmonic identity; the question would be, why? Again, guesswork as I can't see the question, but the difference is between

**time taken**vs**at what time is the second revolution complete**(This paper is horrible)

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#4

(Original post by

THanks for your response, but I'm still confused on one bit- even if I started at say 270 degrees, time for 2 complete revolutions would still be the same as if I started at 180 degrees, right?

**Mazza2000**)THanks for your response, but I'm still confused on one bit- even if I started at say 270 degrees, time for 2 complete revolutions would still be the same as if I started at 180 degrees, right?

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