# Proving indentity..

• February 21st 2010, 07:06 AM
snigdha
Proving indentity..
sin^6 A + cos^6 A= 1 - 3sin^2 A.cos^2 A
• February 21st 2010, 07:42 AM
Soroban
Hello, snigdha!

Quote:

Prove: . $\sin^6\!A + \cos^6\!A\;=\; 1 - 3\sin^2\!A\cos^2\!A$
The left side is a sum of cubes.

$\text{Factor: }\;\underbrace{(\sin^2\!A + \cos^2\!A)}_{\text{This is 1}}\:(\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)$

. . . . . . $=\;\;\sin^4\!A \;-\; \sin^2\!A\cos^2\!A \;+\; \cos^4\!A \;\;=\;\;\sin^4\!A + \cos^4\!A - \sin^2\!A\cos^2\!A$

Add and subtract $2\sin^2\!A\cos^2\!A$

. . . $\underbrace{\sin^4\!A {\color{red}\;+\; 2\sin^2\!A\cos^2\!A} + \cos^4\!A} \;-\; \underbrace{\sin^2\!A\cos^2\!A {\color{red}\;-\; 2\sin^2\!A\cos^2\!A}}$

. . $= \qquad\underbrace{(\sin^2\!A + \cos^2\!A)^2}_{\text{This is 1}} \qquad\quad -\qquad\quad 3\sin^2\!A\cos^2\!A$

. . . . . . . . . . . . . $= \;\;1 - 3\sin^2\!A\cos^2\!A$

• February 21st 2010, 08:00 AM
snigdha
Quote:

Originally Posted by Soroban
Hello, snigdha!

The left side is a sum of cubes.

$\text{Factor: }\;\underbrace{(\sin^2\!A + \cos^2\!A)}_{\text{This is 1}}\:(\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)$

. . . . . . $=\;\;\sin^4\!A \;-\; \sin^2\!A\cos^2\!A \;+\; \cos^4\!A \;\;=\;\;\sin^4\!A + \cos^4\!A - \sin^2\!A\cos^2\!A$

Add and subtract $2\sin^2\!A\cos^2\!A$

. . . $\underbrace{\sin^4\!A {\color{red}\;+\; 2\sin^2\!A\cos^2\!A} + \cos^4\!A} \;-\; \underbrace{\sin^2\!A\cos^2\!A {\color{red}\;-\; 2\sin^2\!A\cos^2\!A}}$

. . $= \qquad\underbrace{(\sin^2\!A + \cos^2\!A)^2}_{\text{This is 1}} \qquad\quad -\qquad\quad 3\sin^2\!A\cos^2\!A$

. . . . . . . . . . . . . $= \;\;1 - 3\sin^2\!A\cos^2\!A$

i got a better clue...
we know that a^3+b^3 = (a+b)^3 - 3ab(a+b)

so by this i proceeded with the problem...& got it proved...
thanks anyway...!!