3sinA + 5cosA = 5 then 3cosA - 5sinA = ?

( a sin A + b cos A )^2 + ( b sin A - a cos A) ^2

= a^2 sin ^2 A + b^2 cos^2 A + 2ab sin A cos A + b^2 sin ^2 A + a^2 cos ^2 A - 2ab cos A sin A
= a^2 + b^2

so
(3sinA + 5cosA)^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so 5^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so (5 sin A - 3 cos A) = +/- 3

or 3 cos A - 5 sin A = +/- 3

Let x and y be positive real numbers. Prove that x^7 + y^7 >= x^4*y^3 + x^3*y^4

x^7 + y^7 - (x^4*y^3 + x^3*y^4)
= x^4(x^3-y^3) - y^4(x^3 - y^3)
= (x^4-y^4)(x^3-y^3)
= (x^2-y^2)(x^2+y^2) (x-y)(x^2 + xy + y^2)
= (x-y) (x+ y)(x^2+y^2) (x-y)(x^2 + xy + y^2)
= (x-y)^2 (x+ y)(x^2+y^2) (x^2 + xy + y^2)

as each term is non negative so
x^7 + y^7 - (x^4*y^3 + x^3*y^4) >= 0 ( > if x and y are not same and = if same)

or x^7 + y^7 >= x^4*y^3 + x^3*y^4

Prove that 1^k+2^k+3^k+...+n^k is devisible by n(n+1)/2 where k is any odd positive integer & n>1 or n=1?

as k is odd we have (a^k+b^k) is divisible by (a+b)

now

n^k + 1 ^k is divisible by (n+1)
(n-1)^k + 2^k is divisible by (n+1)
if n is odd
((n+1)/2)^k is divisible (n+1)/2 so sum is divisible by (n+1)/2

if n is even then there are n/2 terms each is divisible by (n+1)

we consider 2 cases

case 1 n odd :
----------------------
the sum is divisible by (n+1)/2

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divisible by n(as n-1 is even ) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n

so it is divisible by n(n+1)/2 as they are co-primes.

case 2 n be even
------------------------------
the sum is divisible by (n+1)

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divsible by n/2 (as n-1 is odd) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n/2

so it is divisible by n(n+1)/2

so in both cases is is divisible by n(n+1)/2

I am sure some one like will come with a better solution

if n is even we have the sum divisible by n+ 1


if n is odd the middle term is ((n+1)/2)^k

so sum is divisible by (n+1)/2 both cases

prove that the value of (sin x*cos 3x)/(cos x*sin 3x) cannot lie between 1/3 and 3.

it is tan x/ tan 3x

= tan x/ (tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ])
= (1 -3 tan ^2 x)/(3 - tan^2 x )

say tan x = y which can take any value and

t = (1-3y^2)/(3-y^2)

or 3t - ty^2 = 1- 3y^2

or (3-t)y^2 + 3t - 1 = 0

or (3-t)^2y^2 = (3t-1)(t-3)

as LHS >=0 so RHS >=0

(3t-1)(t-3) >= 0 => t >= 1/3 and t >=3 => t >= 3

or t <= 1/3 and t <=3 => t <= 1/3

so cannot lie between 1/3 and 3.

If a and b are the roots of x^2+x+1=0, what is the equation with roots a^19 and b^7?

a^2 = - (a + 1)

so a^3 = -(a^2 + a) = 1

so a^19 = a

similiary

b^7 = b

so roots are a^19 and b^7=> a and b are roots so eqution is x^2+x+1= 0

find x^4 + y^4 + z^4 when

x+y+z=3
x^2+y^2+z^2=5
x^3+y^3+z^3=7

solution

because there are 3 variables let them be solution of

f(t) = t^3 + at^2 + bt + c = 0

sum of roots = 3 so a = - 3

b = (xy + yz + zx) = 1/2((x+y+z)^2 - (x^2+y^2+z^2)) = 2

t^3 - 3t^2 + 2t +c = 0

f(x) = x^3 - 3x^2 + 2x + c = 0
f(y) = y^3 - 3y^2 + 2y + c = 0
f(z) = z^3 - 3z^2 + 2 z + c = 0

so add to get (x^3 + y^3 + z^3) - 3(x^2 + y^2+ z^2) + 2 (x+y+z) + 3c = 0

or 7 - 3 * 5 + 2 * 3 + 3c = 0

or 3c = 2

so f(t) = t^3 - 3t^2 + 2t + 2/3 = 0

multiply by t to get

t^4 - 3 t^3 + 2t^2 + 2/3 t = 0

x,y ,z satisfy this and adding putting x, y, z for t and adding we get

x^4 + y^4 + z^4 = 3(x^3+y^3+z^3) - 2 (x^2 + y^2 + z^2 ) - 2/3(x+y+z)
= 3 * 7 - 2 * 5 - 2/3 * 3 = 21-10 - 2 = 9

(x^2+y^2)/2 where x and y have the same parity. will it result a sum of 2 perfect squares

in other words

x^2 + y^2 =2 (a^2 +b^2) and x^2 + y^2 is even

the solution becomes obvious if you realize that

(a+b)^2 + (a-b)^2 = 2(a^2+b^2)

so x = a+b and y = a- b is a solution and both a and b should be even or odd

Does sqrt(-2i) = 1-i or i-1

both 1-i or i-1 when squared give - 2i

so both are square roots

but when we need to compute sqrt(-2i) this is the principal sqrt and and as per wiki the principal sqrt is defined to be

if z = r e^it then z^(1/2) = r^(1/2)e^(it/2)

-2i = 2 e^i(- pi/2) (note that that angle should be between (-pi to pi] that is pi is inculded and not -pi

so sqrt(-2i) = sqrt(2) e^(-pi/4)i = sqrt(2)( cos (-pi/4) +i sin (-pi/4)) = 1 -i

The no. of roots of equation x^2+x+3+2sinx=0,x belongs to (-180 to +180) is?

x^2+x+3+2sinx
= ( x^2+ x + 1/4)+ 11/4 + 2 sin x
= (x+ 1/2)^2 + 11/4 + 2 sin x

the lowest value of (x+ 1/2)^2 = 0
lowest value of 2 sin x = - 2

so lowest value = 11/4 -2 = 3/4

so it can never be zero

so ans is zero

simplify Sin π/9 sin 2π/9 sin π/3 sin 4π/9

we know sin sin π/3 = √(3)/2

2 Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = Sin π/9 sin π/3 ( 2 sin 2π/9 sin 4π/9)

= Sin π/9 sin π/3 ( cos 2π/9 – 2 cos 2π/ 3) using 2 sin A sin B = cos |(A-B)| - cos (A+B)

= Sin π/9 in π/3 ( Sin π/9 cos 2π/9 + 1/ 2 )

So 4 Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = = Sin π/9 in π/3 ( 2 Sin π/9 cos 2π/9 + 1)

= sin π/3 ( 2Sin π/9 ( 1- 2 sin ^2 π/9) +1 Sin π/9)

= sin π/3 ( 3 Sin π/9 - 4 sin ^3 π/9)
= sin π/3 sin π/3 using sin 3x = 3 sin x – 4 sin ^ 3 x
= sin^2 π/3 = 3/4

Or Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = 3/16