C1 and C2 DOUBTS HERE!!!!!

[**RoRo**]

This is a question from C2, June 2005, it's Question 8.

The circle C, with centre at the point A, has equation x²+y²-10x+9=0.
Find
(a) the coordinates of A, [2 marks]
(b) the radius of C [2 marks]

Can someone please explain to me how do you arrive to the following answers?
(a) x=5, y=0
(b) radius = 4

[narnia]

Solve for 0<=x<360,
cos3x=-1/2

iv done this but dont really know if my answers are correct...
i got 40, 80, 160, 200, 280 and 320 =\


please need the confirmation urgently =)

[elemis]

Solve for 0<=x<360,
cos3x=-1/2

iv done this but dont really know if my answers are correct...
i got 40, 80, 160, 200, 280 and 320 =\


please need the confirmation urgently =)

If you re-insert your answers back into the original equation you should get -0.5.

Try it and see.

[narnia]

If you re-insert your answers back into the original equation you should get -0.5.

Try it and see.

yea thts right on cal...
buh then the answers given are 240 and 120 only...after multiplying x=40 and x=80 by 3...
what about the rest values i got? :S

[narnia]

solve for 2sin(theta)tan(theta)=3 in the range 0<theta<360

[astarmathsandphysics]

here

[narnia]

here

thank you =)

[narnia]

Q1(a) Prove that sinX +sinY=[2sin((X+Y)/2)][cos((X-Y)/2)]
where X=A+B and Y=A-B.

(b)HEnce or otherwise, solve for 0<=theta<360
sin40 + sin20 =0

iv done the (a) part....buh gettin confused for (b).. =\


Q2. 2sin²(theta) -2sin(theta)=cos²(theta)    in the range -180<=theta<180.

have done this too and got +-160.5, +-19.5 and +-90...
buh few of them are wrong when i apply it to the equation....please can someone do the whole of this question and tell me? :S

[astarmathsandphysics]

a)sin (A+B)=sinA cos B+cos A sin B
sin (A-B)=sinA cos B-cos A sin B
Add these two equations and put X=A+B, Y=A-B
b)Put A=3 theta B=theta to get
sin 4 theta +sin 2 theta =2 sin3theta cos theta=0
sin sin 3theta =0 or cos theta =0
3theta =0,180,360,540, 720,900,1080 or theta =0,180,360
so theta =0,60,120,180,240,300,360

2. 2sin^2 x -2 sin =cos^2 x =1-sin^2 x so 3sin^2 x -2sin x -1=0 so (3sin x+1)(sinx -1)=
3sinx+1 =0 so x =-sin^-1 (1/3)=-19.5, 180--19.5=199.5, -180+19.5=-160.5,360+-19.5=340.5
or sinx=1 so x=90

[The Golden Girl =D]

can someone Please do the following graphs in detail -cuz I don't know how to get them -__-
theta is between 360 and o degrees in BOTH these Qs !

y=3 sin (x + 60) - 2     

y = 4 cos3X - 2

I know such a Question won't come in the official exam but it will in MY coming test T_T

Thank you in advance

[elemis]

can someone Please do the following graphs in detail -cuz I don't know how to get them -__-
theta is between 360 and o degrees in BOTH these Qs !

y=3 sin (x + 60) - 2     

y = 4 cos3X - 2

I know such a Question won't come in the official exam but it will in MY coming test T_T

Thank you in advance

[The Golden Girl =D]

graph

Thank you Ari but I want the Steps of How to do it bot what it'll look like 

[elemis]

Thank you Ari but I want the Steps of How to do it bot what it'll look like 

Draw the normal sine curve. Then move it 60 degrees to the left horizontally.

Then multiply the Y coordinates by 3 followed by a vertical translation of 2 units downwards.

[The Golden Girl =D]

I NEED this URGENTLY !!

f(x) = 2x³ + 3x² -6x + 1

a} ...
B] i ) ....
b]ii) Solve equation , f(x) = 0   ---> How do I do that , can you show me in Steps

Thanks in advance =]

[elemis]

I NEED this URGENTLY !!

f(x) = 2x³ + 3x² -6x + 1

a} ...
B] i ) ....
b]ii) Solve equation , f(x) = 0   ---> How do I do that , can you show me in Steps

Thanks in advance =]

Plug in values for x, start methodically from zero until you get a number. In this case I got x=1 as a solution.

Hence, (x-1) is a solution.

Perform algebraic division of the original equation with (x-1) and find the resultant quadratic equation.

Factorise the quadratic.

State the solutions.