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December 11th, 2018, 10:20 AM   #1
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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)
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December 11th, 2018, 10:54 AM   #2
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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

$B = \sin^4(x)-\cos^4(x) - 2\sin^2(x)$
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December 11th, 2018, 11:00 AM   #3
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$\displaystyle \begin{align*}A &= \sin^6(x) + \cos^6(x) + 3\sin^2(x)\cos^2(x) \\
&= \sin^6(x) + \cos^6(x) + 3\sin^2(x)\cos^2(x)(\sin^2(x) + \cos^2(x)) \\
&= (\sin^2(x) + \cos^2(x))^3 \\
&= 1\end{align*}$
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December 11th, 2018, 11:15 AM   #4
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Quote:
Originally Posted by romsek View Post
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.
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December 11th, 2018, 11:49 AM   #5
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$\displaystyle \begin{align*}B &=2\sin^4(x) - \cos^4(x) - 2\sin^2(x) \\
&= \sin^4(x) + (\sin^2(x) + \cos^2(x))(\sin^2(x) - \cos^2(x)) - 2\sin^2(x) \\
&= \sin^4(x) + \sin^2(x) - (1 - \sin^2(x)) - 2\sin^2(x) \\
&= \sin^4(x) - 1 \end{align*}$
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