O-Level Additional Math

[Xiao Hu]

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

[FrekiWang]

Xiao Hu:

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

sqrt[2+sqrt(2+2cosx)]
=sqrt{2+sqrt[2+2(2cos^2(x/2)-1)]}
=sqrt[2+sqrt(4cos^2(x/2))]
Note here, 135<x/2<180 (2nd), so cos(x/2) is negative.
=sqrt[2-2cos(x/2)]<--- I suppose your mistake is because you have +2cos(x/2) in this step, I will explain this at the end.
=sqrt[2-2(1-2(sin^2(x/4))]
=sqrt[4sin^2(x/4)]
Note here, 67.5<x/4<90 (1st), so sin(x/4) is positive.
=2sin(x/4)

So the common mistake here is that some students will assume sqrt(x^2)=x, which is not true.

In fact, sqrt(x^2)=|x|, which is equal to x if x is positive and is equal to -x if x is negative. (e.g. if x = -2, we have sqrt(x^2)=2 which is -x).

Therefore, whenever you need to pull a 'perfect square' from the square root, make sure you determine whether you are pulling the square of a positive or negative number. In this question, cos(x/2) is negative, that is why when you pull cos^(x/2) out of the square root, you have to add a negative sign in front.

Hope you could understand. Cheers

[Xiao Hu]

Hi FrekiWang,

Perfect explanation, you found my mistake. Yes, I undestand that. I think this is a very common pitfall.
Very happy to learn from you, it’s a great site to have you and others help us parents with maths questions that we can’t solve.

Thanks again!
Xiao Hu

[Xiao Hu]

Hi FrekiWang,


Need help on this question, please see the attached picture.

Ans in txtbook:2y=x+4

I just couldn't figure out how to get at least 1 more point on the chord in order to find the equation of the chord. Even with the centre and the radius of the circle given, and the mid-point given, I just couldn't figure it out.

http://i52.tinypic.com/2q0j7ty.jpg\">

(Can't add the picture, it kept prompting can't be more than 640 pixel wide. So I tried breaking it up into 3 lines, hope it's readable.)
Thanks in advance,
Xiao Hu

[FrekiWang]

I suppose you can understand my explanation below:


The line connecting the centre of circle and the midpoint of the chord, is perpendicular to the chord.

You have the coordinates of the centre and the midpoint, thus the gradient of the line. Then you can find the gradient of the chord.

Gradient of line + coordinates of one point will be enough for you to find the equation:D

[Xiao Hu]

Hi FrekiWang,

Oh mine! You are great!! When I first wrote my question, I was saying I just coundn’t find at least 1 more point or a gradient. But I have to rewrite bec was having trouble to load the image of the question. I left out the word gradient.

THanks so much for your prompt reply and kind help!!

You are good,
Xiao Hu

[FrekiWang]

Xiao Hu:

Hi FrekiWang,

Oh mine! You are great!! When I first wrote my question, I was saying I just coundn't find at least 1 more point or a gradient. But I have to rewrite bec was having trouble to load the image of the question. I left out the word gradient.

THanks so much for your prompt reply and kind help!!

You are good,
Xiao Hu

lol, you are welcome

[Xiao Hu]

Hi FrekiWang,


May I ask more about this question I posted.
What if the sign in the question is changed from + to -?

Quote part of your solution below:
=sqrt[2+sqrt(4cos^2(x/2))]
Note here, 135<x/2<180 (2nd), so cos(x/2) is negative.
=sqrt[2-2cos(x/2)]<--- I suppose your mistake is because you have +2cos(x/2) in this step, I will explain this at the end.

So we would have to change the sign to + when we take out the perfect square cos(x/2) out of the root?

Thanks in advance,
Xiao Hu

[FrekiWang]

Whenever you take out the square of a negative expression out the root, you have to add a negative in front.


eg.

sqrt(9a)=3sqrt(a) (you are taking the square of 3 out. so no need to do anything)

Given a>0
sqrt(2a^2)=asqrt(2) (you are taking the square of a out, and you know a is positive, so no need to do anything)

Given a<0
sqrt(2a^2)=-asqrt(2) (you are taking the square of a out, and you know a is negative, so a negative sign is needed)

[Xiao Hu]

CoffeeCat:

midnightspark:

Hi,


Just asking some sec 2 maths questions, please help


1)
A rectangle of sides x cm and y cm has an area of 72 cm2. Another rectangle of sides (x+1.5) and (y-4)cm has the same area. Find the values of x and y.

2)
When 5 is added to both the numerator and denominator of a fraction, the result becomes 1/2. When 1 is subtracted from both the numerator and the denominator, the fraction becomes 1/5. Find the fraction.

3)
The scale of map X is 1: x and the scale of map Y is 1:y. If the same distance is represented as 5cm on map X and 7.5cm on map Y, calculate the ratio x:y.

Thanks.

For qns 2, use algebra.
let the original fraction be x/y.
(x+5)/(y+5) = 1/2
cross multiplying, 2x + 10 = y+ 5
2x + 5 = y
(x-1)/ (y-1) = 1/5
5x - 5 = y - 1
......

For qns 1
xy = 72
(x+1.5)(y-4) = 72
let y= (72/x) and substitute . You will get a quadratic equation after multiplying the whole equation by x.

For qns 3,
5x = 7.5y
x/y = 7.5/5 = ...

hmmm too busy to give you full solutions. hope this helps.

Hi CoffeeCat,
Can you help double check your answer for question 3?
Since for the same distance, it's represented shorter at 5cm than 7.5cm on X vs Y map. So isn't the ration the other way round 5/7.5=2/3?
Hope you wouldn't mind to help clarify and to point out where my mistake in reasoning is.

Thanks,
Xiao Hu.