Solve Sin(^6)x + Cos(^6)x Factoring Problem

  • Thread starter Thread starter riddlingminion
  • Start date Start date
  • Tags Tags
    Factoring
AI Thread Summary
The discussion focuses on factoring the expression sin^6(x) + cos^6(x) using the identity x^3 + y^3 = (x + y)(x^2 - xy + y^2). The participants explore the derivation leading to the conclusion that sin^6(x) + cos^6(x) simplifies to cos^2(2x). There is some confusion regarding the steps, particularly about the terms involved in the factorization. Humor is introduced with a comment about an incorrect equality, highlighting the complexity of the problem. The conversation emphasizes the importance of careful algebraic manipulation in trigonometric identities.
riddlingminion
Messages
7
Reaction score
0
How do I factor sin(^6)x + cos(^6)x?
 
Physics news on Phys.org
Use x^3+y^3=(x+y)(x^2-xy-y^2).
 
Better

\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.

Daniel.
 
Doesn't ( \sin^{2} x +\cos^{2} x ) ( \sin^{4}x +\cos^{4} x ) - 2 \sin^{2}x \cos^{2} x ( \sin^{2} x +\cos^{2} x) = \sin^6 x+ \cos^6 x - \sin^2 x \cos^4 x- \sin^4 x \cos^2 x?:confused:
 
Last edited by a moderator:
dextercioby said:
Better

\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.

Daniel.

Whereby the hitherto unknown equality:
\frac{1}{8}+\frac{1}{8}=0
is proven. :smile:
 
Yes, thought it was 2 simple 2 be true. You can drop that 2, then. :d

Daniel.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Factoring difference of squares not working?
Replies
3
Views
2K
Solve the simultaneous equations involving x, y, and their squares
Replies
2
Views
1K
Solve ##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##
Replies
24
Views
2K
Find the magnitude and the direction of the resultant vector
Replies
8
Views
3K
Solve for ##x## in the Given Problem
Replies
17
Views
1K
Graph Sin(2x+6): Is Parent Function the Best Option?
Replies
6
Views
1K
What Are the Two Factors of 15120 Given Their Sum and Difference?
Replies
5
Views
2K
Solving a trigonometric equation with multiples of ##\tan##
Replies
7
Views
2K
Solve the given trigonometry problem
Replies
7
Views
2K
Find an equivalent equation involving trig functions
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top