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December 11th, 2018, 10:20 AM  #1 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  Show that A and B are constant
Any help with that if you have some free time ? Show that A and B are constant A = sin6(x) + cos6(x) + 3sin2(x).cos2(x) B= 2sin4(x)  cos4(x)  2sin2(x) 
December 11th, 2018, 10:54 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 
I'm assuming this means $A = \sin^6(x) + \cos^6(x) + 3\sin^2(x)\cos^2(x)$ $B = 2\sin^4(x)\cos^4(x)  2\sin^2(x)$ as written $A+B = \sin^4(x)$ is not constant. did you mean $B = \sin^4(x)\cos^4(x)  2\sin^2(x)$ 
December 11th, 2018, 11:00 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
$\displaystyle \begin{align*}A &= \sin^6(x) + \cos^6(x) + 3\sin^2(x)\cos^2(x) \\ &= \sin^6(x) + \cos^6(x) + 3\sin^2(x)\cos^2(x)(\sin^2(x) + \cos^2(x)) \\ &= (\sin^2(x) + \cos^2(x))^3 \\ &= 1\end{align*}$ 
December 11th, 2018, 11:15 AM  #4 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  No, I wrote it the right way, and I should drag out the possibility that it's wrong.
Last edited by skipjack; December 11th, 2018 at 11:50 AM. 
December 11th, 2018, 11:49 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
$\displaystyle \begin{align*}B &=2\sin^4(x)  \cos^4(x)  2\sin^2(x) \\ &= \sin^4(x) + (\sin^2(x) + \cos^2(x))(\sin^2(x)  \cos^2(x))  2\sin^2(x) \\ &= \sin^4(x) + \sin^2(x)  (1  \sin^2(x))  2\sin^2(x) \\ &= \sin^4(x)  1 \end{align*}$ 

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