prove that tan20+4sin20=root 3

tan20+4sin20
= sin 20/ cos 20 + 4 sin 20
= ( sin 20 + 4 sin 20 cos 20) / cos 20
= (sin 20 + 2 sin 40)/ cos 20
= (sin 20 + 2 sin (60-20))/ cos 20
= ( sin 20 + 2 sin 60 cos 20 - 2 cos 60 sin 20)/ cos 20
= (sin 20+ 2 sin 60 cos 20 - sin 20)/ cos 20
= 2 sin 60cos 20/ cos 20
= 2 sin 60
= 2(√3/2) = √3

alternatively


we can proceed from (sin 20 + 2 sin 40)/ cos 20

as ( sin 20 + sin 40 + sin 40) / cos 20
= (2 sin 30 cos 10 + sin 40)/ cos 20
= (cos 10 + sin 40)/ cos 20
= ( sin 80 + sin 40)/ cos 20
= 2 sin 60 cos 20/ cos 20
= 2 sin 60
= √3

sum log(base4)2-log(base8)2+log(base 16) 2 - ..

the nth term = log( base (2^(n+1) 2 = 1/(n+1)

sign is alternating

so we get

S = 1/2 - 1/3 + 1/4 - 1/5 + ....

we have from http://www.mathkb.com/Uwe/Forum.aspx/mat…

ln 2= 1 - 1/2 + 1/3 - 1/4 ....

so ln 2 = 1- s

so S = 1 - ln 2

solve for value of X and Y if X2 +Y2=25 and X3+Y3=91?

Because we are having x^2+y^2 and x^3 + y^3 we can choose

(x+y) = a and xy = b

We get x^2+ y^2 =a^2 -2b = 25 ..1

And (x^3+y^3) = (x+y)^3 – 3xy(x+y)

or a^3 – 3ab = 91 ..2

so 2 (a^3 -3ab) – 3a(a^2- 2b) = 182-75a

or a^3 -75a + 182 = 0

this is cubic in a and by taking factors of 182(1,2,7,13,14,26,91,182) and –ve of them

we can see that a= 7 is a root and from (1) b = xy = 12

now x+ y = 7 and xy = 12 => (x-y)^2 = (x+y)^2 – 4xy = 1

so x- y = +/- 1

x-y = 1 => x= 4 and y = 3

x-y - = - 1 => x = 3 ,y = 4

so solution = (4,3) and (3,4)

find the solutions of tan x=x have a solution

x= 0 is a solution.

for each n there is exactly one solution in the range npi - pi/2 and npi+pi/2 as in this range tan x increases from - infinite to + infinite and at one point it cuts the line y = x

Find all positive integers n such that the decimal representation of n^2 consists of odd digits only

if n is even it is not possible

so n is odd

now (10k+m)^2 = 100k^2 + 20kb + m^2

m = 1 => 100k^2 + 20k + 1: k > 1 means tens digit even so k = 1 => n = 1
m= 3 => 100k^2 + 60k + 9 : k > 1 means tens digit even so k = 0 => n = 3
m = 5 => 100k^2 + 100k + 25 tens digit even
m = 7 => 100k^2 + 140k + 49 => tens digit even
m = 9 => 100k ^2 + 180k + 81 => tens digit even

so only choices n =1 and 3 whose square 1 digit numbers 1 and 9

Note: I could have used (10k+ 5), (10k+/-1) , (10k+/- 3) and could have got result with less computation

If a,b,c are the roots of x^3+4x+1=0 ,then the equation whose roots are a^2/(b+c) ,b^2/(a+c),c^2/(a+b) ?

a,b,c are roots

so we get comparing coefficients of x^2 ( that is sum of roots)

a+ b + c = 0 ..1
now
a^2/(b+c) ,b^2/(a+c),c^2/(b+a) are root of new equation

a^2/(b+c) = a^2/(-a) from (1)

so new equation has roots -a , -b , and -c

f(x) = x^3+4x + 1 = 0 has roots a,b,c

so f(-x) = - x^3 - 4x + 1 = 0 or x^3 + 4x -1 =0 has roots -a , -b , and -c ora^2/(b+c) ,b^2/(a+c),c^2/(b+a)

Show that there are infinitely many basic Pythagorean triples (x,y,z) in which z-x=2

z = (x+2)
if z (and x) even then y is also even so not basic

so z and x both are odd
let z = (2m+1) and x = (2m-1)

(2m+1)^2 - (2m-1)^2 = y^2

or 8m = y^2

so m= y^2/8

y ^2 = 16n^2 shall be a solution as y^2 is divisible by 8 and perfect square so divisible by 16

so m = 2n^2 and there is no other solution

so x = (4n^2-1), y = 4n, z = (4n^2+1) are solutions

there are infinite solutions and lowest 5 are below

n =1 gives (3,4,5)
n= 2 gives (15,8, 17)
n= 3 gives ( 35, 12, 37)
n = 4 gives (63,16,65)
n =5 gives (99,20,101)

What is value of f(x+3) , if f(x+1)=2x^2 -11x+3

we have 
x+ 3 = (x+1) + 2

as f(x+1) = 2x^2 -11x+3

so f(x+3) = 2(x+2)^2 - 11(x+ 2) + 3 = 2(x^2 + 4x + 4) - 11(x+2) + 3 = 2x^2 -3x - 11
----------------------------------------------------------------------------------------------
(rationale

f(x+1) = 2x^2 - 11x + 3

let x + 1 = y

so f(y) = 2(y-1) ^2 - 11y + 3

so f(y+2) = 2(y+1)^2 - 11(y+1) + 3

hence f(x+3) = 2(x+2)^2 - 11(x+2) + 3)

If abc is a three digit number and a, b, c are distinct digits, what is abc if acb+bca+bac+cab+cba=3194

we have (In digit form)
abc = 100a + 10b + c
acb = 100 a + 10c + b
bac = 100b + 10a + c
bca = 100b + 10c + a
cab = 100c + 10 a + b
cba = 100c + 10 b + a

so sum = 222 ( a+ b+ c) > 3194 and 3194 + abc = 222(a+b+c)


find the lower and upper limit of a+b+ c

 a + b + c > 3194/222 = 14.38
so lower limit = 15 and let us start and lowe limit and go one by one finding abc and checking

if a + b+ c = 15 then abc = 222*15-3194 = 136 does not satisfy
a + b + c = 16 , abc = 106 + 222 = 358 yes
a+ b+ c = 17, abc = 358+ 222 = 780 no
a+ b+ c = 18, abc= 780+ 222 > 1000 = 772 no and a+b+ c > 18 is not possible

so abc = 358

Show that 2+i is a root of z^3 - 5z^2 + 9z - 5=0. Find the other 2 roots

one can put z = 2 + i and expand it but it shall be long and prone to error

you can see that it has rational coefficients so if z = 2 + i is a root then z = 2 -i must be a root

and so (z-(2+i))(z-(2 -i) ) or (z-2)^2 +1 or z^2-4z + 5 must devide z^3 - 5z^2 + 9z - 5

by division you see z^3 - 5z^2 + 9z - 5 = (z^2-4z + 5)(z-1)

so z^2-4z + 5 devides z^3 - 5z^2 + 9z - 5

so z = 2 + i is a root
other roots are 2 -i  and 1

Let x & y be positive real numbers such that x^3 + y^3 + 1/27 = xy. Find the value of 1/x.

x^3+y^3+1/27 = xy
=> x^3 + y^3 + (1/3)^3 = 3 x y (1/3)
if a^3 + b^3 + c^3 = 3abc then a + b + c = 0 or a=b= c

as (a^3+b^3+c^3) -3abc = 1/2(a+b+c) ((a-b)^2 + (b-c)^2 + (c-a)^2) 


so x + y + 1/3 = 0 or x = y= 1/3


as x+y+1/3 cannot be zero for x and y positive so x= y = 1/3 

or 1/x = 3