Finding exact value of tan22.5 & 2 polynomial questions (1 Viewer)

[coeyz]

from (1-cos2x)/(1+cos2x) = tan^2x
show that the exact value of tan22.5 is square root 2 -1


8x^4 + 12x^3 - 30x^2 + 17x - 3 =0
Find all the roots.
wht is the quickest way to find roots?
I always try one by one, negative/positive..
it take quite a long time and sometimes still couldnt get the ans =(


If a,B,Y are the roots of 2x^3 - 4x^2 -3x - 1 =0
Find the value of a^2B^2 + a^2Y^2 + B^2Y^2

THANKS A LOT

 

[x3.eddayyeeee]

from (1-cos2x)/(1+cos2x) = tan^2x
show that the exact value of tan22.5 is square root 2 -1


8x^4 + 12x^3 - 30x^2 + 17x - 3 =0
Find all the roots.
wht is the quickest way to find roots?
I always try one by one, negative/positive..
it take quite a long time and sometimes still couldnt get the ans =(


If a,B,Y are the roots of 2x^3 - 4x^2 -3x - 1 =0
Find the value of a^2B^2 + a^2Y^2 + B^2Y^2

THANKS A LOT

FOR SECOND QUESTION.
use the factor/ remainder theroem of polynomials.
that is ;
if (x - a) is a factor , then P(a) = 0
and
for (x- a) ; the remainder when dividing P(x) is P(a)

use those two results to find first 2 or so factors. then use these factor to divide original polynomial.

PS. I tested (x + 3 ) . and thats a factor work from there?

EDIT. are you sure you have typed equation correctly? outside (x+3) can't find anymore.

 

[tommykins]

from (1-cos2x)/(1+cos2x) = tan^2x
show that the exact value of tan22.5 is square root 2 -1

If a,B,Y are the roots of
Find the value of a^2B^2 + a^2Y^2 + B^2Y^2

THANKS A LOT

 

[kurt.physics]

from
show that the exact value of tan22.5 is

THANKS A LOT

We know that from a right angles isosceles triangle with sides 1 - 1 - root 2.

Hence,



multiplying top and bottom by ,



rationalising the denominator,





Hence,

 

[shaon0]

from (1-cos2x)/(1+cos2x) = tan^2x
show that the exact value of tan22.5 is square root 2 -1


8x^4 + 12x^3 - 30x^2 + 17x - 3 =0
Find all the roots.
wht is the quickest way to find roots?
I always try one by one, negative/positive..
it take quite a long time and sometimes still couldnt get the ans =(


If a,B,Y are the roots of 2x^3 - 4x^2 -3x - 1 =0
Find the value of a^2B^2 + a^2Y^2 + B^2Y^2

THANKS A LOT

for 1:
(1-cos2x)/(1+cos2x) = tan^2x
Let x=22.5
Thus, (1-cos(45))/((1+cos(45))=[tan(22.5)]^2
((2-sqrt(2))/(2+sqrt(2))=[tan(22.5)]^2
(2-sqrt(2))^2/2=[tan(22.5)]^2
Thus, sqrt(2)-1=tan(22.5)

 

[tommykins]

Lol the question just got done 3 times.

 

[coeyz]

FOR SECOND QUESTION.
use the factor/ remainder theroem of polynomials.
that is ;
if (x - a) is a factor , then P(a) = 0
and
for (x- a) ; the remainder when dividing P(x) is P(a)

use those two results to find first 2 or so factors. then use these factor to divide original polynomial.

PS. I tested (x + 3 ) . and thats a factor work from there?

EDIT. are you sure you have typed equation correctly? outside (x+3) can't find anymore.

yee thankss
but i mean, how do u find the First root in a quick way? not trying one by one.?
becoz sometimes the root maybe a fraction..

 

[tommykins]

yee thankss
but i mean, how do u find the First root in a quick way? not trying one by one.?
becoz sometimes the root maybe a fraction..

Test the obvious integers such as 1,2,3,4 and then test obvious fractions (1/2, 1/3)

they won't give you any more complex a polynomial

 

[coeyz]

thankyou everyone

 

[tommykins]

for 1:
(1-cos2x)/(1+cos2x) = tan^2x
Let x=22.5
Thus, (1-cos(45))/((1+cos(45))=[tan(22.5)]^2
((2-sqrt(2))/(2+sqrt(2))=[tan(22.5)]^2
(2-sqrt(2))^2/2=[tan(22.5)]^2
Thus, sqrt(2)-1=tan(22.5)

get off your fucking high horse, sif neg rep me for 'too brief an explanation' - wtf

 

[kurt.physics]

^^ yeh i reckon! i got sent neg rep to for 'too long.' (in reference to my work-out in this thread)

WTF is your problem shaon0! Who died and made you prince of Mathematics!

 

[kurt.physics]

get off your fucking high horse...

+1