*"Find, correct to one decimal place, the value of x, 180 ≤ x ≤ 270, which satisfies the equation:*

3cos(2x-40) - 1 = 0 "

3cos(2x-40) - 1 = 0 "

The answer given is

**164.8**, not the same as mine.

Could someone confirm this for me?

Thanks

Ally

Started by Ally, Apr 14 2004 06:13 PM

32 replies to this topic

Posted 14 April 2004 - 06:13 PM

Here is a question about Trigonometry (quite simple really!), but my answer does not agree with the answer given.

*"Find, correct to one decimal place, the value of x, 180 ≤ x ≤ 270, which satisfies the equation:*

3cos(2x-40) - 1 = 0 "

The answer given is**164.8**, not the same as mine.

Could someone confirm this for me?

Thanks

Ally

3cos(2x-40) - 1 = 0 "

The answer given is

Could someone confirm this for me?

Thanks

Ally

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Posted 14 April 2004 - 06:21 PM

Here is another question, about Circles, which I would be grateful if anyone could answer it:

*"Find the equations of the tangents to: x² + y² - 14x - 2y + 5 = 0, which pass through Q(0,2)."*

This time I only come up with 1 equation (wrong aswell) , how can you get two equations for this?

Ally

This time I only come up with 1 equation (wrong aswell) , how can you get two equations for this?

Ally

**Edited by ally, 15 April 2004 - 09:14 AM.**

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Posted 14 April 2004 - 07:02 PM

cos 2x - 40 = 1/3

2x - 40 = 70.5

70.5 is what a calculator gives, but other solutions are 289.5, 430.5, 649.5... etc.

2x = 110.5, 329.5, 470.5, 689.5... etc

x = 55.3, 164.8, 235.3, 344.8... etc.

The only answer between 180 and 270 is 253.3.

164.8 isn't a correct answer because it's not between 180 and 270 !

2x - 40 = 70.5

70.5 is what a calculator gives, but other solutions are 289.5, 430.5, 649.5... etc.

2x = 110.5, 329.5, 470.5, 689.5... etc

x = 55.3, 164.8, 235.3, 344.8... etc.

The only answer between 180 and 270 is 253.3.

164.8 isn't a correct answer because it's not between 180 and 270 !

Posted 14 April 2004 - 10:34 PM

Quote

"Find the equations of the tangents to: x² + y² - 28x - 20y + 5 = 0, which pass through Q(0,2)."

Is the equation and point definitely right?

Posted 15 April 2004 - 12:24 AM

I've worked throught the trig one, and agree with andy.

I also got stuck working through the circles one. I sure hope it was a typo

I also got stuck working through the circles one. I sure hope it was a typo

Posted 15 April 2004 - 09:06 AM

Yes, I've copied it exactly from what is said in the book. *(No I didn't)*

The answers they've given for the circles question is:

** y = 2x + 2 ;**

y = -5.5x + 2

Mind you the book, 'Revise For Higher Mathematics' Heinemann (and the Heinemann textbook we use in school), is hardly a reliable source.

Ally

The answers they've given for the circles question is:

y = -5.5x + 2

Mind you the book, 'Revise For Higher Mathematics' Heinemann (and the Heinemann textbook we use in school), is hardly a reliable source.

Ally

**Edited by ally, 15 April 2004 - 09:16 AM.**

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Posted 15 April 2004 - 09:11 AM

Woops!

I have copied it wrong.

I will edit the above post.

Sorry!

(I Still don't get the same answer though)

I have copied it wrong.

I will edit the above post.

Sorry!

(I Still don't get the same answer though)

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Posted 15 April 2004 - 10:52 AM

Ally, I worked through that question and my answer agrees with the textbook.

The method I used was:

I've uploaded my working in case you want to see.

The method I used was:

- Sub the point into y-b=m(x-a) to get the form of the equations
- Sub in the resulting y = ... into the circle equation to get points of intersection
- The result is a quadratic with unknown coefficients. For tangency, b² - 4ac = 0
- Solve the quadratic to get values for m
- Sub these into the y = ... to get the equations of the lines

I've uploaded my working in case you want to see.

Posted 15 April 2004 - 06:37 PM

Wo I sure hope I dont get a question like that in my exam. It was so confusing. Will we get questions like that in the exam?? I was confident now I'm all worried?? AHHH!

Half ideas,half quality, half a million pound law suit!

Posted 15 April 2004 - 06:51 PM

I think that question is very much level A, but it's not *too* bad if you know how to tackle it.

It's quite similar to standard questions about intersections between a line and a circle, the twist being that you are normally given*m* and asked for the point.

I'm sure that even if you never got all the marks, you'd manage to pick up some if you had a stab at it. The main thing is bagging all the easy marks, so don't worry about the odd tricky question

It's quite similar to standard questions about intersections between a line and a circle, the twist being that you are normally given

I'm sure that even if you never got all the marks, you'd manage to pick up some if you had a stab at it. The main thing is bagging all the easy marks, so don't worry about the odd tricky question

Posted 15 April 2004 - 07:04 PM

Cheers!

Thanks a lot for that, it was really helpful.

You must be a genious!

I would never have thought of it as that way.

What I done is that I :

Found out the centre of the circle - C - (7,1)

Then used both C and Q to find the gradient.

I then found the perperdicular gradient.

I then substituted Q and gradient into y-b=m(x-a)

I think that would actually give the equation of the tangent to the curve,**assumming** that the point Q lies on the circle.

Ally

(it was quite a tough 1)

Thanks a lot for that, it was really helpful.

You must be a genious!

I would never have thought of it as that way.

What I done is that I :

Found out the centre of the circle - C - (7,1)

Then used both C and Q to find the gradient.

I then found the perperdicular gradient.

I then substituted Q and gradient into y-b=m(x-a)

I think that would actually give the equation of the tangent to the curve,

Ally

(it was quite a tough 1)

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Posted 15 April 2004 - 07:15 PM

Thanx George. Yeh I did it the same way as you Ally. Good to know your way of doing though George. It'll help me in Advanced Higher.

Half ideas,half quality, half a million pound law suit!

Posted 15 April 2004 - 08:32 PM

Here is another question to test your brains, I'm having difficulty with this 1 aswell (not doing to well am I! lol)

It's a Trig Question :

*If f(a) = 6sin²a - cos a, express f(a) in the form pcos²a + qcos a + r. Hence solve, correct to 3 decimal places, the equation 6sin²a - cos = 5*

And yes I have copied it down correctly this time lol.

Ally

It's a Trig Question :

And yes I have copied it down correctly this time lol.

Ally

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Posted 15 April 2004 - 08:41 PM

Taking the first bit of the question...

6sin²a - cosa

sin²a can be rewritten as (1 - cos²a)

this is because the trig rule sin²a + cos²a = 1, rearranging that gives sin²a = 1 - cos²a

so that leads you to

6(1 - cos²a) - cos a

multiply out the bracket

6 - 6cos²a - cos a

rearrange

-6cos²a - cos a + 6

so thats in the form the question asks for (p = -6, q = -1, r = 6)

Then the next part of the question:

using the answer from above you can get

-6cos²a - cos a + 6 = 5

-6cos²a - cos a + 1 = 0

using the quadratic formula:

-b±√(b² - 4ac)/2a

substitue the values into the formula (a = (-6), b = (-1), c = 1)

you get 1±√(25)/-12

giving you the two answers -1/2 and 1/3

so cos a = -1/2 and 1/3

solving those equation**in radians** leads you to the answer 1.231 and 2.094

6sin²a - cosa

sin²a can be rewritten as (1 - cos²a)

this is because the trig rule sin²a + cos²a = 1, rearranging that gives sin²a = 1 - cos²a

so that leads you to

6(1 - cos²a) - cos a

multiply out the bracket

6 - 6cos²a - cos a

rearrange

-6cos²a - cos a + 6

so thats in the form the question asks for (p = -6, q = -1, r = 6)

Then the next part of the question:

using the answer from above you can get

-6cos²a - cos a + 6 = 5

-6cos²a - cos a + 1 = 0

using the quadratic formula:

-b±√(b² - 4ac)/2a

substitue the values into the formula (a = (-6), b = (-1), c = 1)

you get 1±√(25)/-12

giving you the two answers -1/2 and 1/3

so cos a = -1/2 and 1/3

solving those equation

Posted 16 April 2004 - 11:36 AM

Can anyone who has the Heinemann Maths Textbook, lead me through the steps in obtaining the answer to question 11 on page 67.

Its in the 'Trigonometry:graphs and equations' chapter.

Thanks

Ally

Its in the 'Trigonometry:graphs and equations' chapter.

Thanks

Ally

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Posted 16 April 2004 - 12:36 PM

(a)

If you look at the graph the amplitude has to be 3, since this is half the vertical distance between the maximum and minimum. Since this is a sine graph, the value for p must be**negative** 3 (as the graph is flipped from a normal looking sine graph).

So p = -3

Normally, the graph of y = -3sin(x+r) would have a maximum of 3 and minimum of -3. But in this case its maximum is 4 and minimum -2, so the graph has been moved up 1 place.

So q = 1

If you think again of a normal inverted sine graph, it will start at 0, with a minimum turning point at 90º, but in this graph it is at 50º. Therefore the graph has been moved to the left 40º. Moving the graph to the left means the value of r is positive 40.

So r = 40

Then for u, once again think of the normal inverted sine graph which would have a maximum turning point at 270º. Since we're moving the graph to the left by 40º, this turning point will now be at 230º.

So u = 230

( b )

For the value s, this cuts the y axis. Therefore x must be 0.

The values from above can be subsituted into the equation to give:

y = -3 sin (x + 40) + 1

Making x 0 leads to y = -3 sin (0 + 40) + 1

Working that out on the calculator gives - 0.928

So s = -0.928

For t, this cuts the x-axis. So y must be 0.

- 3 sin (x + 40) + 1 = 0

Rearrange giving

sin (x + 40) = 1/3

Inverse of sin gives 19.47

Using quadrants 1 and 2 you get

x + 40 = 19.47, 160.53

x = -20.53, 120.53

So t must equal 120.5

I know those answers don't match the book but they match the past paper answers. The book's answers are right they've just taken the graph at a different angle, as though it's a normal sine curve that's been moved along to the right before beginning.

If you look at the graph the amplitude has to be 3, since this is half the vertical distance between the maximum and minimum. Since this is a sine graph, the value for p must be

So p = -3

Normally, the graph of y = -3sin(x+r) would have a maximum of 3 and minimum of -3. But in this case its maximum is 4 and minimum -2, so the graph has been moved up 1 place.

So q = 1

If you think again of a normal inverted sine graph, it will start at 0, with a minimum turning point at 90º, but in this graph it is at 50º. Therefore the graph has been moved to the left 40º. Moving the graph to the left means the value of r is positive 40.

So r = 40

Then for u, once again think of the normal inverted sine graph which would have a maximum turning point at 270º. Since we're moving the graph to the left by 40º, this turning point will now be at 230º.

So u = 230

( b )

For the value s, this cuts the y axis. Therefore x must be 0.

The values from above can be subsituted into the equation to give:

y = -3 sin (x + 40) + 1

Making x 0 leads to y = -3 sin (0 + 40) + 1

Working that out on the calculator gives - 0.928

So s = -0.928

For t, this cuts the x-axis. So y must be 0.

- 3 sin (x + 40) + 1 = 0

Rearrange giving

sin (x + 40) = 1/3

Inverse of sin gives 19.47

Using quadrants 1 and 2 you get

x + 40 = 19.47, 160.53

x = -20.53, 120.53

So t must equal 120.5

I know those answers don't match the book but they match the past paper answers. The book's answers are right they've just taken the graph at a different angle, as though it's a normal sine curve that's been moved along to the right before beginning.

Posted 16 April 2004 - 02:56 PM

Thanks a lot for that!!!

It really helped.

Ally

It really helped.

Ally

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Posted 16 April 2004 - 04:07 PM

No problem

Posted 17 April 2004 - 03:11 PM

Ok I have a problem from Maths in Action book page55.

It has a pic of the graph y=2cos2x -1 for {0 lessthan/equal to x less than equal to pi}

(A) solve an equation to find the x co-ordinates of A and B

{ A and B are where the graph crosses the x -axis}

so I take it you ho 2cos2x - 1 = 0 and solve it, I don't get the correct answers.

(B) Find the x co-ordinates of the points of intersection of the curve and the line y= -2

So I take it you solve 2cos2x - 1 = -2 but again I don't get the rite answers.

If anyone could work these out I'd be grateful!

It has a pic of the graph y=2cos2x -1 for {0 lessthan/equal to x less than equal to pi}

(A) solve an equation to find the x co-ordinates of A and B

{ A and B are where the graph crosses the x -axis}

so I take it you ho 2cos2x - 1 = 0 and solve it, I don't get the correct answers.

(B) Find the x co-ordinates of the points of intersection of the curve and the line y= -2

So I take it you solve 2cos2x - 1 = -2 but again I don't get the rite answers.

If anyone could work these out I'd be grateful!

**Edited by sparky, 17 April 2004 - 03:11 PM.**

Mark

Posted 17 April 2004 - 03:17 PM

2cos2x - 1 = 0

2 cos 2x = 1

cos 2x = 0.5

2x = 60, 300, 420, 660, etc.

x = 30, 150 = pi/6, 5pi/6

2 cos 2x = 1

cos 2x = 0.5

2x = 60, 300, 420, 660, etc.

x = 30, 150 = pi/6, 5pi/6

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