[Next Level]

Hey guys, stuck on q.8(ii)(b) on this paper:

http://pdf.ocr.org.uk/download/pp_10..._gce_4723.pdf?

So you need to use part 8(i), the answer to which is root 18 cos(x-0.785).

So for 8(ii)(b), here is my working:

8/(root 18 cos(x-0.785) ) = 8/9 root 6
therefore 8 = (root 18 cos (x-0.785).8 root 6) / 9
therefore 8 = (48 root 3 cos (x-0.785) ) / 9 = 8
therefore 48 root 3 cos(x-0.785) = 72
therefore cos(x-0.785) = (root 3) / 2
hence 3x - 0.785 = 1/6 pi
therefore x = 5/36 pi (this is what the expression simplifies to).

However, the answer is 1/36 pi, no idea where I've gone wrong...any ideas?

Pos rep for a clear explanation Cheers!

[TenOfThem]

instead of pi/6 use -pi/6 which has the same cos

[Next Level]

(Original post by TenOfThem)
instead of pi/6 use -pi/6 which has the same cos

ahh I see, how did u quickly realise that the cos of pi/6 is the same as the cos of -pi/6?

[TenOfThem]

(Original post by Next Level)
ahh I see, how did u quickly realise that the cos of pi/6 is the same as the cos of -pi/6?

CAST or graphs

whatever method you normally use to find the "other" solutions

For me it is CAST

[Next Level]

(Original post by TenOfThem)
CAST or graphs

whatever method you normally use to find the "other" solutions

For me it is CAST

ahh right usually for cos I'd just do 360 - (the other value), but thinking about a cast diagram the negative value of the first solution is the equivalent of that no?

[raheem94]

(Original post by Next Level)
ahh I see, how did u quickly realise that the cos of pi/6 is the same as the cos of -pi/6?

If you have studied C2, then you should be knowing that

[TenOfThem]

(Original post by Next Level)
ahh right usually for cos I'd just do 360 - (the other value), but thinking about a cast diagram the negative value of the first solution is the equivalent of that no?

Yes

[NutterFrutter]

(Original post by Next Level)
ahh right usually for cos I'd just do 360 - (the other value), but thinking about a cast diagram the negative value of the first solution is the equivalent of that no?

For something like this it'd be better to be able to picture the cosine graph, there's perfect symmetry. As raheem94 has said cos(x) = cos(-x).

Remember for sine, sin(-x) = -sin(x)

[TenOfThem]

(Original post by raheem94)
If you have studied C2, then you should be knowing that

TBH you would have learnt this at GCSE

[raheem94]

(Original post by TenOfThem)
TBH you would have learnt this at GCSE

Is this taught at GCSE?

I didn't knew about this until i did C2.

[Next Level]

Cheers guys!

[TenOfThem]

(Original post by raheem94)
Is this taught at GCSE?

I didn't knew about this until i did C2.

yes

and the trig graphs ... using them to find other values

and sin^2 + cos^2 = 1

and sin/cos = tan

and the question you referred to earlier where you are given cos of an angle and expected to find sin or tan using the triangle

[NutterFrutter]

(Original post by TenOfThem)
yes

and the trig graphs ... using them to find other values

and sin^2 + cos^2 = 1

and sin/cos = tan

and the question you referred to earlier where you are given cos of an angle and expected to find sin or tan using the triangle

This is on the Edexcel GCSE specification? I can't remember ever studying this at GCSE.

[raheem94]

(Original post by TenOfThem)
yes

and the trig graphs ... using them to find other values

and sin^2 + cos^2 = 1

and sin/cos = tan

and the question you referred to earlier where you are given cos of an angle and expected to find sin or tan using the triangle

GCSE is quite different to the GCE 'O' Level which i did.

My one didn't included anything from the list except the last point of using that triangle though i do believe that overall it was much tougher than GCSE. But as it was very old spec, so didn't had any good books to self-study it, and the grade boundaries were pathetic when i took the exam.

[TenOfThem]

(Original post by NutterFrutter)
This is on the Edexcel GCSE specification? I can't remember ever studying this at GCSE.

Trig graphs and the trig values of other angles so knowing cos(a) = cos(-a) is certainly there

When I referenced the others I was referring to questions that have been asked on AQA that expect students to use pythagoras and trig on the non-calculator paper