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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: 2,574 Thanks: 1420 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 $B = \sin^4(x)-\cos^4(x) - 2\sin^2(x)$ Thanks from topsquark
 December 11th, 2018, 11:00 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,019 Thanks: 2253 \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*} Thanks from SkyCod
December 11th, 2018, 11:15 AM   #4
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Joined: Dec 2018
From: Canada

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Math Focus: mainly algebra
Quote:
 Originally Posted by romsek 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 $B = \sin^4(x)-\cos^4(x) - 2\sin^2(x)$
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: 21,019 Thanks: 2253 \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*} Thanks from topsquark and SkyCod

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