Core 3 Maths - Double Angle Question - Am I Right?

Squarehed

Squarehed

Squarehed

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Hi guys, just wanting to check I'm right in my method here as I don't have the mark scheme for this practice paper to hand.

Question 8(a) of June 2006 C3 paper:

Given that cos A= 3/4, where 270°< A <360°, find the exact value of sin 2A

I know double angle formula: sin2A = 2sinAcosA

My working was like this:

If cosA = 3/4. Pythagoras gives us a triangle of √7, 3, 4

Therefore sinA = √7/4

Putting these into sin2A = 2sinAcosA gives: 2(√7/4)(3/4)

This gives us the value of sin2A as: (3√7/8)

Have I gone about this the right way or is it wrong to assume that I can use pythagoras?
 
EDIT: Wait a minute

EDIT2: I am pretty sure you are correct.

EDIT3: OK. I can see why you might have doubted your answer here due to the 270 < A < 360 part of the question but the answer is the same due to the related angles. But you knew that anyway! As beenom says, anything involving exact values is almost certainly going to be a surd.
 
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Quite correct and a method I suspect most of us wouldn't have even thought of. Too much calculator use for us. We've forgotten the basics.
 
Yep, always remember that if they say find the exact value of such and such, you have to use pythagoras' theorem. Using the calculator would provide the angle A in decimals which is not what you want here. :)
 
Core 3. Easy stuff. ;)

If you want a really high grade, then do every exam paper you can get ahold of, it helps loads.
 
Oh yeah, past papers are where its at. I'm trying to keep my average up from last year w/ core 1,2 and s1 coming in at 98/100 UMS pts average. Don't feel as confident for this one, but its Thursday afternoon so will have hopefully got sorted by then.

As for this question, I had used my calculator at first but remembered halfway through that it'd probably round the irrational values into decimals. I am right in thinking I can always assume we have a right angled triangle for this type of question?

Thanks for the help :)
 
Ugh, I'm getting things confused with something we did way back at GCSE. With trig in non-rt angled triangles? Now I'm all embarrassed! I've been out of c3 practice after teaching myself D1 over the hols and revising other subjects :s

Thanks again for your help guys. Night!
 
Actually I'm not sure it's quite right (but it's been a long time :p)

Squarehed said:
Given that cos A= 3/4, where 270°< A <360°, find the exact value of sin 2A
Click to expand...

I think the constraint of 270°< A <360° is important, A is the outside angle of the triangle your working has assumed it's the inside angle (which would be 360°-A).

Squarehed said:
I know double angle formula: sin2A = 2sinAcosA

My working was like this:

If cosA = 3/4. Pythagoras gives us a triangle of √7, 3, 4
Click to expand...
You get lucky here because cos(A) = cos(360° - A) so this still true ;)

Therefore sinA = √7/4
Click to expand...
This is wrong though as sin(A) = -sin(360° - A)

So you've lost a minus sign here.

Putting these into sin2A = 2sinAcosA gives: 2(-√7/4)(3/4)

This gives us the value of sin2A as: -(3√7/8)

Have I gone about this the right way or is it wrong to assume that I can use pythagoras?
Click to expand...

No a fine method just an annoying gotcha ;)
 
Squarehed said:
Ugh, I'm getting things confused with something we did way back at GCSE. With trig in non-rt angled triangles? Now I'm all embarrassed! I've been out of c3 practice after teaching myself D1 over the hols and revising other subjects :s

Thanks again for your help guys. Night!
Click to expand...


a^2=b^2+c^2-2bc CosA


I think, not had to use that in ages mind




Good luck in C3, I had mine on Friday, was much easier than expected.
 
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See what I got wrong there. Is there anyone else who is moved to near-violence by the sound of that man's voice, it drives me insane!
 
I did this just the other day, but in my usual silly ways, forgot the fractions were multiplied so became stuck calculating the answer! I'm doing Edexcel C3 on Thursday afternoon as well, but have run out of the past papers my school has so am having to resort to some Solomon papers which I personally hate! :p
 
g0th2000 said:
Actually I'm not sure it's quite right (but it's been a long time :p)



I think the constraint of 270°< A <360° is important, A is the outside angle of the triangle your working has assumed it's the inside angle (which would be 360°-A).


You get lucky here because cos(A) = cos(360° - A) so this still true ;)


This is wrong though as sin(A) = -sin(360° - A)

So you've lost a minus sign here.



No a fine method just an annoying gotcha ;)
Click to expand...

As usual my rubbish maths skills are exposed. I thought I took related angles into account but as usual if I do stuff in my head I make a mess of it!
 
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