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Maths C3 - Trigonometry... Help??

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Can anyone help me with this?...

C3 Exe 7 Q7e.png
Original post by Philip-flop
Can anyone help me with this?...

C3 Exe 7 Q7e.png


sin(π2)=12 sin(\frac{\pi}{2}) = \frac{1}{2} and cos(π6)=32 cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}
Original post by NotNotBatman
sin(π2)=12 sin(\frac{\pi}{2}) = \frac{1}{2} and cos(π6)=32 cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}


I'm assuming you meant sin(π6)=12 sin (\frac{\pi}{6}) = \frac{1}{2}

Thanks a lot for your help :smile:

How did you realise to do that from that equation? I never would have spotted it if it wasn't for your help :frown:
(edited 7 years ago)
Ffs, now I'm stuck on part (g).
C3 Exe 7 Q7e.png
Am I even doing it right for this question?...
Attachment not found


EDIT: Don't worry, I managed to work this out in a very strange way :tongue:
(edited 7 years ago)
Reply 284
Guys do you know how to solve these questions. They are from C1

1. Find vaalues of the constants A,B and C in identity
4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.
Original post by AB J
Guys do you know how to solve these questions. They are from C1

1. Find vaalues of the constants A,B and C in identity
4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.


Expansion may help :h: you can see what A is going to be anyway, because there's only one x^2 term on the RHS.

For 2, if you are given that a quadratic has equal roots.. what calculation can you perform?
Original post by AB J
Guys do you know how to solve these questions. They are from C1

1. Find vaalues of the constants A,B and C in identity
4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.

Please start a new thread. This is not C3 trigonometry.
Original post by Philip-flop
Ffs, now I'm stuck on part (g).
C3 Exe 7 Q7e.png
Am I even doing it right for this question?...
Attachment not found


EDIT: Don't worry, I managed to work this out in a very strange way :tongue:


Note that cos(90u)=sinu\cos(90-u)= \sin u, then rewrite the second sin term as an appropriate cos.
I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
C3 Exe 7A Q12(a).png
(edited 7 years ago)
Original post by Philip-flop
I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
C3 Exe 7A Q12(a).png


I would change the sin^2(theta) into 1 - cos^2(theta) and then work from there
Original post by asinghj
I would change the sin^2(theta) into 1 - cos^2(theta) and then work from there


oh right so I would do?...

cos2θsin2θ cos^2 \theta - sin^2 \theta

cos2θ(1cos2θ) cos^2 \theta - (1-cos^2 \theta)

cos2θ1+cos2θ cos^2 \theta - 1 + cos^2 \theta

2cos2θ1 2cos^2 \theta -1.... But then where do I go from here? :frown:
Original post by Philip-flop
oh right so I would do?...

cos2θsin2θ cos^2 \theta - sin^2 \theta

cos2θ(1cos2θ) cos^2 \theta - (1-cos^2 \theta)

cos2θ1+cos2θ cos^2 \theta - 1 + cos^2 \theta

2cos2θ1 2cos^2 \theta -1.... But then where do I go from here? :frown:


what is the double angle formula:
cos2θ cos2 \theta ?
Original post by asinghj
what is the double angle formula:
cos2θ cos2 \theta ?

I'm not sure I know. The Edexcel C3 Modular Maths textbook hasn't even explained double angle formulas. But from what I've just found out...

cos2θ cos2 \theta

=cos(θ+θ) = cos( \theta + \theta)

=cosθcosθsinθsinθ = cos \theta cos \theta - sin \theta sin \theta ... and so on from there (what I showed in my last comment).
Original post by Philip-flop
I'm not sure I know. The Edexcel C3 Modular Maths textbook hasn't even explained double angle formulas. But from what I've just found out...

cos2θ cos2 \theta

=cos(θ+θ) = cos( \theta + \theta)

=cosθcosθsinθsinθ = cos \theta cos \theta - sin \theta sin \theta ... and so on from there (what I showed in my last comment).


oh ok.. so yes

=cosθcosθsinθsinθ = cos \theta cos \theta - sin \theta sin \theta

this can be simplified to

cos2θsin2θ cos^2 \theta - sin^2 \theta

because cosθcosθ=cosθ×cosθ cos \theta cos \theta = cos \theta \times cos \theta
and this made me realise you didn't have to change the sin2θ sin^2 \theta into anything...

oh and watch this video
http://www.examsolutions.net/tutorials/identities-for-sin2a-cos2a-tan2a/?level=A-Level&board=Edexcel&module=C3&topic=1420
(edited 7 years ago)
Original post by asinghj
oh ok.. so yes

=cosθcosθsinθsinθ = cos \theta cos \theta - sin \theta sin \theta

this can be simplified to

cos2θsin2θ cos^2 \theta - sin^2 \theta

because cosθcosθ=cosθ×cosθ cos \theta cos \theta = cos \theta \times cos \theta
and this made me realise you didn't have to change the sin2θ sin^2 \theta into anything...

oh and watch this video
http://www.examsolutions.net/tutorials/identities-for-sin2a-cos2a-tan2a/?level=A-Level&board=Edexcel&module=C3&topic=1420

Oh right, I see. So for this question my answer would literally be...

cos2θsin2θ cos^2 \theta - sin^2 \theta

=cos2θ =cos 2\theta
Original post by Philip-flop
I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
C3 Exe 7A Q12(a).png

For part a), later on in the textbook you'll learn about the double angle formula and the expression will be a form you recognise.

At this stage of the textbook, you should be looking to write the expression as the right-hand-side as one of the addition formulae.

So cos2θsin2θ=cosθcosθsinθsinθ\cos^2 \theta - \sin^2 \theta = \cos \theta \cos \theta - \sin \theta \sin \theta.

Can you see what to do from here?
Original post by notnek
For part a), later on in the textbook you'll learn about the double angle formula and the expression will be a form you recognise.

At this stage of the textbook, you should be looking to write the expression as the right-hand-side as one of the addition formulae.

So cos2θsin2θ=cosθcosθsinθsinθ\cos^2 \theta - \sin^2 \theta = \cos \theta \cos \theta - \sin \theta \sin \theta.

Can you see what to do from here?

I don't think I know where to go from there.

All I know is...

cosθcosθsinθsinθ cos \theta cos \theta - sin \theta sin \theta

=cos(θ+θ) = cos( \theta + \theta)

=cos2θ = cos2 \theta
Original post by Philip-flop
I don't think I know where to go from there.

All I know is...

cosθcosθsinθsinθ cos \theta cos \theta - sin \theta sin \theta

=cos(θ+θ) = cos( \theta + \theta)

=cos2θ = cos2 \theta

You don't need to do anything else. cos2θ\cos 2\theta is a single trig function so you're done.
Original post by notnek
You don't need to do anything else. cos2θ\cos 2\theta is a single trig function so you're done.


Oh I see. Thank you so much @notnek !! :smile:
Ok. How do I even do part (c) and (d)?? :frown:

C3 Exe 7A Q12(a).png

For part (c)... Shall I start off by using the fact that...
tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos \theta} ?
(edited 7 years ago)

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