• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
112
Reaction score
24
Points
28
guys i need your help.....
P,Q and R are the angles of a triangle such that cosP=3/5 and cosQ=5/13
Without using a calculator find the value of cosR [2]
 
Messages
1,243
Reaction score
1,891
Points
173
guys i need your help.....
P,Q and R are the angles of a triangle such that cosP=3/5 and cosQ=5/13
Without using a calculator find the value of cosR [2]
R=180-(P+Q) cos(R)=cos(180-(P+Q))= - cos( P+Q) (from identity cosx=-cos(180-x)) now simply find -cos(P+Q) by using the formula cos(A+B)=cosAcosB-sinAsinB
and sinx=(1-(cosx)^2)^(1/2)
 
Messages
112
Reaction score
24
Points
28
I got another question......
the variable complex number z is given by
z=2cosA+i(1-2sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............
 
Last edited:
Messages
1,243
Reaction score
1,891
Points
173
I got another question......
the variable complex number z is given by
z=2cosA+i(1-sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............
are u sure the question is right ?
 
Messages
8,477
Reaction score
34,837
Points
698
I got another question......
the variable complex number z is given by
z=2cosA+i(1-sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............
1/4?
Substitute that z in that denominator so it will be :
1/(2cos(theta) - i(sinn(theta)))
then solve this as u usually solve your complex number problems when u have denominator in terms of i.
you will get 1/4 as the real part. :)
 
Last edited:
Messages
237
Reaction score
266
Points
73
Picture 030.jpg
okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?
 
Messages
112
Reaction score
24
Points
28
View attachment 60930
okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?
substitute any value when x<-7 we get a positive value which is not the solution (>0)
substitute any value when x>1 we get a positive value which is not the solution (>0)
sustitute values when -7<x<1 we get a negative value which is the solution....[<0]
 
Messages
8,477
Reaction score
34,837
Points
698
View attachment 60930
okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?
They have used that to show the points satisfying in that region. The same is explained by :
substitute any value when x<-7 we get a positive value which is not the solution (>0)
substitute any value when x>1 we get a positive value which is not the solution (>0)
sustitute values when -7<x<1 we get a negative value which is the solution....[<0]

The easy way to remember this thing is :
if any inequality with < 0
then the roots will be sth like : less positive/ more negative value < x < more positive/ less negative value

if with > o
then, x > more positive/ less negative value and x < less positive/ more negative value

lets take ur example : x = 1 and x = -7 and inquality < 0
thius the answer will be -7 < x < 1 :D :cool:
 
Messages
112
Reaction score
24
Points
28
They have used that to show the points satisfying in that region. The same is explained by :


The easy way to remember this thing is :
if any inequality with < 0
then the roots will be sth like : less positive/ more negative value < x < more positive/ less negative value

if with > o
then, x > more positive/ less negative value and x < less positive/ more negative value

lets take ur example : x = 1 and x = -7 and inquality < 0
thius the answer will be -7 < x < 1 :D :cool:

you can also memorise this
if (x-a)(x-b)<0 then a<x<b if a<b
b<x<a if a>b
if (x-a)(x-b)>0 then x<a , x>b if a>b
x>a ,x<b if a<b
 
Messages
2,206
Reaction score
2,824
Points
273
Yep that's an A for sure
ARE U SURE I WILL GET AN A?

Recently one of my friend said that total percentage of A levels is divided like this :
30% P1 and 20% S1 = 50% As level
30% P3 and 20% S2 = 50% A2 level
So overall A level grade will be depended on both, so as I scored 70% (b) in As level now I will have to get 90% (a) in my A2 level exam to score overall (A).
Is it like this or how did u said YES that's A for sure? :O Please temme asap, i wanna know how much hard work i should do xD
 
Top