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Posted July 23rd, 2009 by alpha
in

If A+B+C=$ \pi $
show that $ sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}=1+4sin\frac{(\pi-A)}{4}sin\frac{(\pi-B)}{4}sin\frac{(\pi-C)}{4} $

‹ Prove that triangle is either an isosceles or a right angled triangle. Show that the the triangle is isosceles ›

Comments

answer

On July 24th, 2009 Structure says:
$$ sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}-1=2sin\frac{A+B}{4}\cos \frac{A-B}{4}-(1-\cos\frac{A+B}{2})=$$
$$2sin\frac{A+B}{4}\cos \frac{A-B}{4}-2\sin^2\frac{A+B}{4}=2\sin\frac{A+B}{4}(\cos \frac{A-B}{4}-\sin\frac{A+B}{4})$$
$$=4sin\frac{(\pi-A)}{4}sin\frac{(\pi-B)}{4}sin\frac{(\pi-C)}{4} $$

~~THANK YOU ~~

On July 26th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

A query

On August 7th, 2009 alpha says:

First of all sorry for posting this under an inappropriate category .
Why am I not allowed to post any more questions?
Please let me know.

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On August 8th, 2009 Structure says:

In the last weeks our site has been attacked by spam and we have to stop this attacks.
You are a trusted user and we shall allow your post as soon as possible.
Best regards

Meantime, I want to post my question here.

On August 9th, 2009 alpha says:

G.P
1)If x,y,z be respectively pth,qth and rth terms of a G.P, prove that (q-r)logx+(r-p)log y+(p-q)logz=0

2)If A.M of two numbers be twice of their G.M , show that the numbers are in the ratio
$ 2+\sqrt{3}:2-\sqrt{3} $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Solution of 2nd question-- please check

On August 9th, 2009 alpha says:

Let the numbers be a and b
$ \frac{a+b}{2}=2\sqrt{ab} $
$ \frac{a+b}{2\sqrt{ab}}=2 $
Using componendo and dividendo
$ \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1} $
$ \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=3 $
$ \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\sqrt{3} $
Again using componendo and dividendo
$ \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} $
$ \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} $
$ \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} $
-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Another question-- a difficult one

On August 9th, 2009 alpha says:

(1+5^-1)(1+5^-2)(1+5^-4)(1+5^-8)............(1+5^-2^n)=5/4(1-5^-2^{n+1})

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On August 9th, 2009 Structure says:
$$(1-x)(1+x)(1+x^2)(1+x^{2^2})(1+x^{2^3})...(1+x^{2^n})=1-x^{2^{n+1}}$$
$$(1+x)(1+x^2)(1+x^{2^2})(1+x^{2^3})...(1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x}$$

Then take

$$x=\frac{1}{5}$$

As

$$(1-x)(1+x)=1-x^2$$
$$(1-x^2)(1+x^2)=1-x^{2^2}$$
$$(1-x^{2^2})(1+x^{2^2})=1-x^{2^3}$$

Complex number

On August 12th, 2009 alpha says:

Find the square roots of the following complex numbers :
a) -i
b) $ x+i\sqrt{x^4+x^2+1 } $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On August 15th, 2009 Structure says:
$$(\frac{\sqrt 2}{2}-i\frac{\sqrt 2}{2})^2=-i$$
$$(-\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2})^2=-i$$
$$(\pm(\sqrt{x^2+x+1}+i\sqrt{x^2-x+1}))^2=x+i\sqrt{x^4+x^2+1}$$

A question on miscellaneous series

On August 23rd, 2009 alpha says:

Find the sum of infinite series
$ 1+(1+a) r +(1+a+a^2)r^2+(1+a+a^2+a^3)r^3+................................. $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On August 23rd, 2009 Structure says:

For a not equal to 1,mulpitpy by (1-a) and then you two geometric series
For a=1 take the derivative of partial geometric series

~~THANK YOU ~~

On August 24th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Another question on series

On August 24th, 2009 alpha says:

$ 1^2 + (1^2+2^2) + (1^2+2^2+3^2)........................ n terms $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

My approach

On August 24th, 2009 alpha says:

nth term of the series would be $ \sum_{1}^{n}k^2=\frac{n(n+1)(2n+1)}{6} $
The sum of the series =$ \sum_{1}^{n}\frac{k(k+1)(2k+1)}{6} =\frac{1}{6}\sum_{1}^{n}{k(k+1)(2k+1)} $
$ =\frac{1}{6} \sum_{1}^{n}2k^3+3k^2+k=\frac{1}{6}[2\sum_{1}^{n}k^3+3\sum_{1}^{n}k^2+\sum_{1}^{n}k] $

Is my procedure correct?

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt
You just have to change general term

$$\sum_{k=1}^{k=n}\sum_{i=1}^{i=k}i^2$$

''ALPHA" =)

ALGEBRA

On September 1st, 2009 alpha says:

If x+y-z=xyz
then find the value of
$ \frac{2x}{1-x^2 }+\frac{2y}{1-y^2}+\frac{\sqrt{z}}{1-y^2} $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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