A2 Level Maths Trig Question

Hi,
I am going to attatch some workings I have done for a Trig Question. Bearing in mind I have taught myself this topic today so I am a bit rusty. I will also attach the mark scheme so please could someone please explain where I have gone wrong? Thank you so much.

Amelia

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(edited 6 years ago)

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Reply 1

Notnek

21

Original post by aliasanoynmous

Hi,
I am going to attatch some workings I have done for a Trig Question. Bearing in mind I have taught myself this topic today so I am a bit rusty. I will also attach the mark scheme so please could someone please explain where I have gone wrong? Thank you so much.

Amelia

Attachment not found

I think it's mostly okay after a quick read through. But can you explain how you got the

\frac{7}{12}\pi

solution?

Of course you're not using "hence" in this case so your working won't follow the mark scheme but that's fine. The "hence" method is quicker but can be confusing for a lot of students.

Reply 2

aliasanoynmous

OP

8

Original post by Notnek

I think it's mostly okay after a quick read through. But can you explain how you got the

\frac{7}{12}\pi

solution?

Of course you're not using "hence" in this case so your working won't follow the mark scheme but that's fine. The "hence" method is quicker but can be confusing for a lot of students.

Thank you for checking my anwer.I got this answer because when using the cast diagram I was under the impression that if you were doing the inverse of a negative number you had to make it positive when finding your answer. Then, instead of using the "tan" and "all" regions of the diagram you use the "sin" and "cos" regions. Thus, what I have done here. This has usually worked with the other questions and in L6.

Reply 3

Notnek

21

Original post by aliasanoynmous

Thank you for checking my anwer.I got this answer because when using the cast diagram I was under the impression that if you were doing the inverse of a negative number you had to make it positive when finding your answer. Then, instead of using the "tan" and "all" regions of the diagram you use the "sin" and "cos" regions. Thus, what I have done here. This has usually worked with the other questions and in L6.

Where did the

\frac{7}{12}\pi

solution come from : the

= 2 + \sqrt{3}

equation or the

= -2 + \sqrt{3}

equation?

I can't see the

-\frac{1}{12}\pi

solution on any CAST diagram so I'm wondering if you've considered it?

Reply 4

aliasanoynmous

OP

8

Original post by Notnek

Where did the

\frac{7}{12}\pi

solution come from : the

= 2 + \sqrt{3}

equation or the

= -2 + \sqrt{3}

equation?

I can't see the

-\frac{1}{12}\pi

solution on any CAST diagram so I'm wondering if you've considered it?

The CAST diagram is on my second page of workings. No I did not include the (-1)/12 on any CAST diagram because it is negative. The 7/12pi came from the -2+root3. So I did tan^-1 of 2+root3 which gave 5/12pi then I did pi- 5/12pie to get 7/12pi.

(edited 6 years ago)

Reply 5

Notnek

21

Original post by aliasanoynmous

The CAST diagram is on my second page of workings. No I did not include the (-1)/12 on any CAST diagram because it is negative. The 7/12pi came from the -2+root3. So I did tan^-1 of 2+root3 which gave 5/12pi then I did pi- 5/12pie to get 7/12pi.

It seems like you might be missing some understanding of the CAST diagram.

You should have two separate CAST diagrams for the two equations and you shouldn't be mixing up the two which you have done I think.

So for

\tan \theta = 2 + \sqrt{3}

, you use your calculator to get the solution

\theta = \frac{5\pi}{12}

which you did. Then

2+\sqrt{3}

is positive so you're looking for solutions in A and T. The T solution goes beyond the range and you already have the A solution so you're done for this equation i.e. the only solution is

\frac{5\pi}{12}

. Just because the other equation has a negative right-hand-side, this does not affect the first set of solutions.

Then for

\tan \theta = -2 + \sqrt{3}

, you need another CAST diagram and first mark on

\theta = -\frac{\pi}{12}

because this is the calculator solution. Then since

-2+\sqrt{3}

is negative, you need to find soltutions in the S quadrant, since this is the only negative quadrant in the range.

If you're still stuck, please explain which specific part of my post you didn't understand. Explaining/understanding CAST in a forum post can be hard :smile:

Also it's very important that you understand what you're doing when using the CAST diagram and not just copying a method. If you feel you don't understand it fully then I recommend watching videos on the topic.

Reply 6

davros

16

Original post by aliasanoynmous

The CAST diagram is on my second page of workings. No I did not include the (-1)/12 on any CAST diagram because it is negative. The 7/12pi came from the -2+root3. So I did tan^-1 of 2+root3 which gave 5/12pi then I did pi- 5/12pie to get 7/12pi.

More importantly, have you looked at the mark scheme method to understand what they did?

When you see the word "Hence" in a maths question it's usually a pointer to tell you that you can get a particular result fairly efficiently by using previous parts of a question. In this case you've made yourself an awful lot of extra work - and introduced the possibility of errors - by attacking the problem in a roundabout manner :smile:

A2 Level Maths Trig Question

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