solve 2sin(10º)sin(20º + θ) = sin(θ)

we have
2 sin(10º)sin(20º + θ) = 2 sin(10º)( sin20º cos θ+ cos 20º sin θ ) = sin θ

or 2 cos θ sin 10º sin20º
=  sin θ( 1- 2 sin 10º cos 20º)
= 2 sin θ( sin30º - sin 10º cos 20º)
= 2  sin θ(sin(20º + 10º) -  sin 10º cos 20º)
= 2  sin θ(sin 20º cos 10º)
= >  tan θ = tan 10º or θ = 10º

prove that 2^n > n^3 for n > 10

Basic Step:
When n = 10, the inequality 2^n > n^3 is true as 2^10 = 1024 > 1000

Inductive Step:
Assume n = k is true, where k ≥ 10, then
2^k > k^3 ................ (1)

we have for k >= 10

k^3 > 10k^2 > 4k^2 or  3k^2 + k^2 > 3k^2 + 3k + 1

So, from (1)
2(2^k)> 2(k^3)
=>2(2^k) > k^3 + k^3
=>2^(k+1) > k^3 + 3k^2 + 3k + 1
=>2^(k+1) > (k + 1)^3

So n = k + 1 is also true

so induction step is proved and hence proved

 

Starting at the origin, a beam of light hits a mirror (in the form of a straight line) at point A = (4, 8) and is reflected to point B = (8, 12). Compute the exact slope of the mirror.

Slope of the incident ray = (8-0)/(4-0) = 2 ( angle A)

Slope of the reflected ray = (12-8)/(8-4) = 1( angle B)

Now let us find the line that bisects the 2 rays

Tan (A+B) = (tan A + tan B)/(1- Tan A tan B) = (1+2)/(1-2) = - 3

Let C be the angle bisector between A and B

A + B = 2C say m is slope of the bisector

So we get 2m/(1-m^2) = - 3

Or 2m = -3 + 3m^2
Or 3m^2 – 2m – 3 = 0
=> m = (1/6) [2 + √(4 + 36)] ... [m positive taken to the correct slope
=> m = (1/3) [1 + √(10)].

this is the slope of the mirror

The reason for the slope to be positive is that from 4,8 one line goes to (8,12) that is line is pointing towards 1st quadrant and (0,0) is to the direction of 3rd quadrant. so the tangent should have a slope between 1 and 2 and not the normal

Resolve the following expression into partial fraction: X^2/(x+1)^3

This type of problem can be done as

Let x^2/(x + 1)^3 = A/(x + 1) + B/(x + 1)^2 + C/(x + 1)^3
=> x^2 = A(x + 1)^2 + B(x + 1) + C
x = - 1 => C = 1

Comparing coefficients of x^2 and x,
A = 1 and 2A + B = 0 => B = - 2A = - 2
=> x^2/(x + 1)^3 = 1/(x + 1) - 2/(x + 1)^2 + 1/(x + 1)^3.

Which is correct and nothing is wrong in it. This is conventional approach.

However seeing that (x+1)^3 in the denominator it can be done by putting

x+ 1 = t or x= (t-1)
we get x^2/(x+1)^3 = (t-1)^2/t^3 = (t^2 - 2t + 1)/t^3 = 1/t - 2/t^2 + 1/t^3 

             = 1/(x+1) - 2/(x+1)^2 + 1/(x+1)^3

which reduces the number of steps.  

find the smallest positive integer value of n for which {(1+i)^n}/{(1-i)^(n-2)} is a real number ?

((1+i)/(1-i)) = (1+i)^2/2

so ((1+i)/(1-i))^n = (1+i)^(2n)/2^n)

so {(1+i)^n}/{(1-i)^(n-2)} = (1+i)^(2n)/(2^n) (1-i)^2
= ((1+i)^n(1-i))^2 /(2^n)
= (1+i)^(n-1)/2^(n-1)

so (1+i)^(n-1) must be real or n-1 = 0 (by inspection) or n = 1 is the lowest n

if one needs to solve for all n the 1+ i makes 45 degrees with x axis and n = 4t is the solution ( t integer)

solve x^2 - y^2 = 2764 for integer x and y > 0

x^2 - y^2 = (x + y)(x - y) = 2764
 so both (x + y) and (x - y) are factors of 2764.

x+ y and x- y both are even or odd ( as one even and on odd shall  give fractions)

 as product is even so x + y and x-y both are even

Now factorize 2764 into product of primes: 2764 = 4*691 = 2*2*691

so x+ y= 1382 and x- y =2 => x = 692 and y =690 is the only solution

show that x = cos(π/10) is a root of the equation 16x^4 - 20x^2 + 5 = 0.

proof:

let y =  π/10

we have

 cos5y = 16cos^5 y - 20cos^3 y + 5cosy

or 0 =  16cos^5 y - 20cos^3 y + 5cosy

= (16cos^4 y - 20cos^2 y + 5)cos y

as cos  y is not zero 

 so (16cos^4 y - 20cos^2 y + 5) = 0

hence x =  cos(π/10) is a root

Prove that (sinA+sinB)(sinB+sinC)(sinC+sin A) > sin A sinB sinC in a triangle ABC

we have in a triangle a + b > c

so sin A + sin B > sin C ..1

as a/ sin A = b/ sin B = c/ sin C

similarly

( sin B + sin C) > sin A ..2

sin C + sin A > sin B ...3

by multiplying (1) (2) and (3) and knowing that both sides are positive ( sin is positive for any angle of a triangle)

we get the result

(sinA+sinB)(sinB+sinC)(sinC+sinA)>sinA sinB sinC

proved

solve 3 * 4^x - 6^x = 2 * 9^x

we have 4^x = 2 ^ 2x
6^x = 2^x 3^ x
9^x = 3^2x

let 2^x = a and 3^x = b ( we have a and b are not independent of each other) ..1

3a^2 - ab = 2b^2

or 3a^2 - ab - 2b^2 = 0

or (3a + 2b)(a-b) = 0

a = b or (a/b) =1 = (2/3)^x ( from 1) so x = 0

or 3a = - 2b which is not possible as a and b both >0 from (1)

so x = 0

show that for any real a,b both not zero a^2+ab+ b^2 > 0

proof:
      a^2+ab+b^2 = (a^3-b^3)/(a-b)

without loss of generality let a <= b ( as the expression is symetric)

if a < b

then a^3 < b^3 and both a^3-b^3 and a-b are -ve and so ratio is positive

if a = b then a^2 + ab + b^2 = 3a^2 and is zero only if a = 0 that is a=b = 0

proved

solve arcsin(x)+arctan(x) = 0

we note that for x > 0 arc sin x and arctan x both are > 0
for x < 0 arcsin x and arctan x both are < 0

and  at x = 0 both are zero so x = 0 is the solution

Suppose two integers, m and n differ by 1. Suppose also that x, y, and a are integers such that x + a = m^2 and y + a = n^2. Show that xy + a is a perfect square.

without loss of generality as it is symmetric let m = n+ 1 
x=(n+1)^2-a ;
y=n^2 -a

xy + a = ( (n+1)^2 - a)( n^2 - a) +a
= (n^2)(n+1)^2 - (2) (n)(n+1) (a) + a^2
= ((n)(n+1) - a)^2

therefore xy+a is a perfect square.

What is the smallest symmetrical number greater than 56,789 which is exactly divisible by 7?

The number has to be form

10000x+1000y+100z+10y+z
= 10001x+ 1010y + 100z

10001x + 1010y+100z mod 7 = 5x + 2y + 2z mod 7

x cannot be < 5. so let x = 5 and let us look for solution y minimum 6

so 25+2y+2z mod 7 = 0 or 2y+2z = 3 mod 7 or y + z = 5 mod 7

y+z = 5 no solution
so y + z = 12 if y = 6 z = 6 not possible

so y = 7 and z = 5 possible

number = 57575

it is 7 * 8225
if we did not find a solution with x = 5 then we should have tried at x = 6