Mathematics: Post your doubts here!

[Maayee]

Can anyone help me with...
parts ii), iii) and v)
the answers are...
ii) 2(80-x)=160-t^2
iii)0<t<4
v)60m

 

[The Sarcastic Retard]

http://studyguide.pk/Past Papers/CI... AS Level/9709 -Mathematics/9709_w07_qp_2.pdf
q(7)(i)

 

[Maayee]

solving left hand side,
open up the bracket which gives u... cos^2x+6sinxcosx+9sin^2x

use the identity sin2x=2sinxcosx
so u need to get 6sinxcosx so multilpy it by 3... 3sin2x=6cosxsinx

equation becomes cos^2x+3sin2x+9sin^2x
so simplifying it further cos^2x+3sin2x+9(1-cos^2x)
cos^2x+3sin2x+9-9cos^2x
-8cos^2x+9+3sin2x

if u look at 5-4cos2x....its basically the same as -8cos^2x+9 when u simplify it so replace it with that...
thus 5-4cos2x+3sin2x proven

u need to show them how u converted the 5-4cos2x into -8cos^2x+9

 

[holoholo]

Can someone please show me how to do this ?

 

[nehaoscar]

Can someone please sketch the diagram for (b)(i) please?
So I know the first one is a circle with centre 2+i and radius 1
Now the second equation i solved
(x+iy+i)<(x+iy-2)
this gives
y>2x-1.5
So i drew a line like so...
Is it correct as the ms says this :

 

[nehaoscar]

Part ii please

 

[nehaoscar]

View attachment 59189
Can someone please show me how to do this ?

 

[holoholo]

View attachment 59202

How do you find the least value of |z-w| ?

 

[nehaoscar]

How do you find the least value of |z-w| ?

Least value is the distance D-C
so find AC by using sine rule
AC = 3root2/2
DC = AC - 1
Since the radius of circle is 1
So DC = (3root2/2)-1 = 1.12

 

[holoholo]

Can someone please show me how to do this ?

 

[nehaoscar]

View attachment 59232

Can someone please show me how to do this ?

 

[holoholo]

View attachment 59233

Can you please show it in a diagram ?

 

[nehaoscar]

Red is y=x
Blue is y=(1/root3)x +2

 

[holoholo]

View attachment 59234
Red is y=x
Blue is y=(1/root3)x +2

I mean the argand diagram if you can please ?

 

[nehaoscar]

I mean the argand diagram if you can please ?

That is the same as the argand diagram ... y corresponds to imaginary axis and x corresponds to real axis

 

[holoholo]

That is the same as the argand diagram ... y corresponds to imaginary axis and x corresponds to real axis

Can you please show how to obtain the answers using trigonometry on the diagram ?

 

[nehaoscar]

Can you please show how to obtain the answers using trigonometry on the diagram ?

Magnifying the triangle we can find the angles and use the sine rule
The argument of blue line is given in question as pi/6
red is y=x hence the argument is pi/4
Knowing these 2 angles you can find the other angles
We know the blue line cuts the imaginary axis at 2 (when we solved the equation above) so the length will be 2
Now apply the sine rule as x = [2/sin(pi/2)] * [sin(2pi/3)]

 

[Konstantino Nikolas]

acosθ + bsinθ = rsin(θ+α) or rsin(θ-α) or rcos(θ+α) or rcos(θ-α)

How do I know when to use what? :O

 

[nehaoscar]

acosθ + bsinθ = rsin(θ+α) or rsin(θ-α) or rcos(θ+α) or rcos(θ-α)

How do I know when to use what? :O

Learn them:
acosθ + bsinθ = rcos(θ-α)
acosθ - bsinθ = rcos(θ+α)
asinθ + bcosθ = rsin(θ+α)
asinθ - bcosθ = rsin(θ-α)
When cos is first it's rcos and the signs in brackets are opposite to LHS
When sin is first it's rsin and the signs in brackets are same as LHS

 

[Konstantino Nikolas]

Learn them:
acosθ + bsinθ = rcos(θ-α)
acosθ - bsinθ = rcos(θ+α)
asinθ + bcosθ = rsin(θ+α)
asinθ - bcosθ = rsin(θ-α)
When cos is first it's rcos and the signs in brackets are opposite to LHS
When sin is first it's rsin and the signs in brackets are same as LHS

I didn't know this ... damn, tysm!!!