2009/009) Prove that tan (A+B) = (tan A+ tan B)/(1-tan A tan B)

Proof:

We know that

the argument of a complex number z = x + yi is the angle to the real axis

then if we take the 1st number 1 + ai and argument is A and 2nd number 1+bi and argument is B

then a = tan A and b = tan B

1st number = 1 + ai = 1 + i tan A
2nd number = 1+ bi = 1 + i tan B


When we multiply the arguments add so angle is A + B

Now (1+ ai)(1+bi) = (1-ab + i(a+b))

As argument is A+B hence

tan (A+B) = (a+b)/(1-ab)= (tan A + tan B)/(1- tan A tan B) (note tan(arg) = y/x

2009/008) If a,b,c are in AP then prove that (b+c)^2 - a^2 , (c+a)^2 - b^2 , (a+b)^2 - c^2 are in AP

let the terms be a1, a2,a3
a1 = (b+c)^2 - a^2 = (a+b+c)(b+c-a) = (a+b+c)(a+b+c - 2a)
a2 = (c+a)^2 - b^2 = (a+b+c)(a+c-b) = (a+b+c)(a+b+c - 2b)
a3 = (a+b)^2 - c^2 = (a+b+c)(a+b-c) = (a+b+c) (a+b+c - 2c)

as a b c are in ap so are -a , -b , -c so are -2a, -2b, - 2c

addiing constant to them we have

(a+b+c-2a), (a+b+c-2b), (a + b - 2c) are in AP

multiplying by (a+b+c) we have
(a+b+c)(a+b+c-2a), (a+b+c)(a+b+c-2b), (a+b+c) (a + b - 2c) are in AP

hence proved

2009/007) prove sin5A = 5sinA - 20 sin^3A + 16 sin ^5 A

We Know
Sin (A + B) = sin A cos B + cos A sin B
Putting B = 4A we get
sin 5A = sin ( A + 4 A)
= sin A cos 4A + cos A sin 4A
= sin A (1- 2 Sin^2 2A) + cos A 2. Sin 2A cos 2A
( putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = 2A)
= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)
(putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = A)
= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)
= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 Sin ^2 A)
= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^4 A)
= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

(note: edited the above based on 3rd comment below to keep the flow)

2009/006) integers x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.

Let S be set of integers x such that x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.


Proof:

we know
a^3+ b^3+ c^3-3abc = (a+bw+cw^2)(a+bw^2+cw) where w = cube root of -1

let

f(a,b, c) = a+b+c
g(a,b,c) = (a+bw+cw^2)
and h(a,b,c) = (a+bw^2+cw)


then a^3+ b^3+ c^3-3abc = f(a,b,c) g(a,b,c) h(a,b,c)

and x^3+ y^3+ z^3-3xyz = f(x,y,z) g(x,y,z) h(x,y,z)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z)

now let us evaluate f(a,b,c) f(x,y,z) , g(a,b,c) g(x,y,z) and h(a,b,c) h(x,y,z)

f(a,b,c) f(x,y,z) = (a+b+c)(x+y+z) = (ax+ay+ az + bx +by + bz + cx + cy+ cz)

g(a,b,c) g(x,y,z) = (a+bw+cw^2) (x+yw+zw^2)

= (ax +ayw+azw^2 + bxw + byw^2 + bzw^3 + cxw^2 + cyw^3 + czw^4)
= (ax +ayw+azw^2 + bxw + byw^2 + bz + cxw^2 + cyw^3 + czw) knowing w^3 = 1 and hence w^4 = w
= (ax +bz+cy + (ay + bx + cz)w+ (az + by+cx) w^2)
similarly
h(a,b,c) h(x,y,z) = (a+cw+ bw^2) (x+zw+ yw^2)

= (ax+bz+cy) + (az+by+cx)w+ (ay+bx+cz) w^2

If we put ax + bz+ cy = p, ay+bx+cz = q, az+by+cx = r

We get

f(a,b,c) f(x,y,z) = p + q + r = f(p,q,r)

g(a,b,c) g(x,y,z) = p + qw + r w^2 = g(p,q,r)

h(a,b,c) h(x,y,z) = p + rw + q w^2 = h(p,q,r)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z) = f(p,q,r) g(p,q,r) h(p,q,r) = ( p^3+ q^3+ r^3-3pqr)

hence proved

2009/005) parametric form for Pythagorean triplet (Gemoetric way)

To generate canonical form of Pythagorean triplet(geometric way)

Introduction

First it makes sense to define what is Pythagorean Triplet.

A Pythagorean triple is a triple of positive integers a, b, and c such that a right angled triangle exists with legs a , b and hypotenuse c. By the definition of Pythagorean Theorem , this is equivalent to finding positive integers a,b and c satisfying

a2+b2 = c2 ….1

The smallest and best-known Pythagorean triple is a =3, b =4, c = 5

It is not unusual to look for primitive Pythagorean triples.

The (1) is equivalent to finding real solution of

(a/c)2+(b/c)2 = 1
or x2+y2= 1 … 2

In case we get rational solutions in x and y the set (x,y,1) satisfy the condition. We can convert them to integer by proper multiplication that is LCM of denominator.
So we need to find rational x and y to satisfy the equation 2
We can put y as a function of x as following
Y = 2(1-x2)
We can choose x to be a rational number <1 all="" be="" br="" but="" cases.="" cases="" does="" for="" general="" generate="" get="" in="" may="" not="" of="" one="" rational.="" rational="" shall="" some="" the="" this="" to="" x="" y="">Now the problem is can we generate x and y to be rational or integer roots for a b c.

There are a couple of methods to find the rational x y.

the equation x^2+y^2 = 1 is a circle at centre (0,0) and radius 1

If we can find one point in the circle which is a rational point and draw a straight line y= mx +c through that point with m and c to be rational then it shall intersect the circle at another point which has got rational point. By choosing different m we can get more and more rational points.

We know one of the point in the circles is
x = 0 and y = -1 (people have used x = -1 and y =0 but this gives me simpler approach. The choice is arbitrary and no specific choice)
putting in y = mx + c we get y = mx-1 ….3
now put it in the equation 2 to get
x2 + (mx-1)2 = 1
or x2+m2x2-2mx=0
or x(1+m2) -2m = 0
x = 2m/(1+m2) ..4
put x = 2m/(1+m2) in 3 to get
y = 2m2/(1+m2) – 1 or (m2-1)/(m2+1)
so (2m/(1+m2), (m2-1)/(m2+1),1) satisfies x^2+y^2 = 1.
multiplying by m2+1 we get

(2m,.(m2-1),(m2+1)) satisfies x^2+y^2 = z^2 but they are not integers

By choosing different m we get different values basically in parametric form.
Now putting m =u/v which is ratio of integers and multiplying by v2
We get

x = 2uv
y = u2- v2
and z = u2 + v2

so (2uv, (u^2-v2) and (u^2+v^2)) form a triplet

2009/004) Prove the identity sin^2Acos^2B-cos^2Asin^2B = sin^2A-sin^2B

sin^(2)Acos^(2)B-cos^(2)Asin^(2)B
= sin ^2 A(1- sin ^2 B) - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 A sin ^2 B - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 B(sin ^2 A + cos ^2 A)
= sin ^2 A - sin ^2 B

2009/003) factor (x-y)^5+(y-z)^5+(z-x)^5

if a + b+ c = 0

then (a+b)^5 = -c ^5

so a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b + b^5 = - c^5

so a^5 + b^5 +c^5 = - (5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b)
= -5 ab(a^3 + 2 a^2b + 2 a b^2 + b^3)
= - 5 ab((a+b)^3 - (a^2 b + ab^2)
= - 5ab((a+b)3 - ab(a+b))
= - 5ab(a+b)((a+b)^2 - ab)
= 5abc(c^2-ab) as a+b = - c

as (x-y)+ (y-z) + (z-x) = 0

we get 5(x-y)(y-z)(z-x)((z-x)^2 - (x-y)(y-z))
= 5(x-y)(y-z)(z-x)(z^2 + x^2- 2xz -xy + xz +-y^2 -xy)
= 5(x-y)(y-z)(z-x)(x^2+y^2+z^2 - xy - yz - zx)