32 replies to this topic

[Ally]

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

[Ally]

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

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

[AndyW]

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 !

[Steve]

Quote

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

I'm not sure that this question is possible. If you use the distance formula to work out the distance from the centre of the circle at (14, 10) to the point Q(0, 2), the distance turns out to be less than the radius of the circle - which means the point must be inside the circle and a tangent can't pass through it!

Is the equation and point definitely right?

[George]

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

[Ally]

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

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

[Ally]

Woops!

I have copied it wrong.

I will edit the above post.



Sorry!

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

[George]

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

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

Hope that's clear enough!

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

[Discogirl17]

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!

[George]

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

[Ally]

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)

[Discogirl17]

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.

[Ally]

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

[Allan]

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

[Ally]

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

[Allan]

(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.

[Ally]

Thanks a lot for that!!!

It really helped.


Ally

[Allan]

No problem

[sparky]

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!

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

[AndyW]

2cos2x - 1 = 0
2 cos 2x = 1
cos 2x = 0.5
2x = 60, 300, 420, 660, etc.
x = 30, 150 = pi/6, 5pi/6