Circular functions help

[Momo.05]

There is a question that seems straight forward.. that is but when i started to work on it, something went wrong.. so hope someone out there can help me out with it :/

We are asked to solve cos(x-y) when sin(x)=0.6, x is between [pi/2, pi] and tan(y)=2.4, y is between [0, pi/2]

So from the previous questions i was able to solve for ;
cos(x) = -4/5
cos(y) = 5/12 or 0.38

sin(x)= 3/5
sin(y) = 12/13

So using the compound angle formulas i know
cos(x-y) = cos(x)cos(y)+sin(x)sin(y)

So by subbing in the values.. i got this as my equation
cos(x-y) = cos(-4/5)cos(5/12)+sin(3/5)sin(12/13)

So my answer turns out to be 1.00005 .. which is wrong since the B.O.B answer is 16/65 or 0.25

WHAT AM I DOING WRONG ? =,="

[GerrySly]

Whenever given values you just work out corresponding values based on exact triangles



Now using the compound angle formula, we just simplify input those values...

[jimmy999]

You're replacing x and y with the calculated values instead of replacing cos(x)..etc with the values

You should end up with

[Momo.05]

OH.. Is all i can say...
I got it .. i did cos(x) = cos(-0. instead of just -0.8 !

Thanks to all for pointing that out

Now back to maths >:/

[Momo.05]

This is another question about circles, once again im stuck so please help out

Can someone show me how to solve this question

"The points (8,4) and (2,2) are the ends of a diameter of a circle. Find the coordinates of the centre and the radius of the circle."

Thanks you .. I figured i'll re-post here since i didnt want to start too many topics and spam this forum with endless S.O.S on spesh questions posted by me lol

[TrueTears]

Well the midpoint of the line segment with endpoints (8,4) and (2,2) are the centre of the circle.

Further more the length of the line segment is the diameter, so the radius is just half of that.

[Momo.05]

So the answer would be , center = (5,3) and radius of root10 ?
Sorry no answers to check my working out :/

[Momo.05]

I need some help finding the asymptotes of this equation, 4x^2 - 8x - y^2 +2y = 0

I changed the equation into the hyperbola way, which equals ;

4(x-1)^2/3 - (y-1)^2/3 = 1

Sorry if the layout looks confusing, idk how to make it look "math-y" like other people...

I need to find the asymptotes but when i apply the formula with this equation my answer isnt right - multiple of times  >:/

[jimmy999]

First you want to get it into the form of

To do this, you complete the square on both the x and y bits.
You'll eventually get

Apply this to the asymptote equations

You should be able to get the answer from that

[Momo.05]

Urghhh.. im still not doing something right .. can you, when u have time show me your working out for this asymptote of this question please ?

[Momo.05]

Actually, I got it !!.. instead of using 2 i used 4 ..
Took me a while to realise this until i notice the square-ing of my things from your equation - If that made any sense at all

But thank you anyways, and thank for showing me that "LaTex" thing as well :)

[jimmy999]

Good work. Most people put down instead of
It's a very common mistake. I made that mistake when working out the solution

Oh and you will love LaTeX. If you have any assignments throughout the year do them in LaTeX and your teachers will be very impressed

[Momo.05]

LOL.. to be honest lets hope i dont have ANY assignments to impress my teacher throughout the year lol

[Juddinator]

When we have reciprocal circular functions, such as cosex(x), sec(x) and cot(x) and we want to find the y-axis intercept, how do you go about finding it's exact answer as we cannot really use our calculators to give us its exact value. For example if we have sec(pie/3)=2, how do we know it equals two?

The same goes for finding its adjusted asymptotes. How it is done? I seem to have lost my notes... :|

Thanks

[fady_22]

Just change it into a familiar form: e.g.