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April 28th, 2018, 06:39 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 238 Thanks: 1  How $3sin^3 \theta cos^3\theta$ obtained?
$15 (\sin^4X + \cos^4X)  10 (\sin^6X  \cos^6X) = ?$ $\rightarrow 15 (12 \sin^2 \theta. \cos^2\theta)  (13\sin^3 \theta \cos^3\theta)$ $\rightarrow 15  30 \sin^2 \theta. \cos^2\theta  10 + 30 \sin^2 \theta. \cos^2\theta$ $\rightarrow 1510 = 5$ How $3\sin^3 \theta \cos^3\theta$ obtained? Last edited by skipjack; April 28th, 2018 at 08:01 PM. 
April 28th, 2018, 08:16 PM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  
April 28th, 2018, 08:20 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,308 Thanks: 2443 Math Focus: Mainly analysis and algebra 
I suspect that you have a typo. \begin{align*} \sin^6{\theta} + \cos^6{\theta} &= (\sin^2{\theta} + \cos^2{\theta})^3  3\sin^2{\theta}\cos^4{\theta}  3\sin^4{\theta}\cos^2{\theta} \\ &= 1  3\sin^2{\theta}\cos^2{\theta}(1\sin^2{\theta})  3\sin^4{\theta}\cos^2{\theta} \\ &= 1  3\sin^2{\theta}\cos^2{\theta} + 3\sin^4{\theta}\cos^2{\theta}  3\sin^4{\theta}\cos^2{\theta} \\ &= 1  3\sin^2{\theta}\cos^2{\theta} \end{align*} 
April 28th, 2018, 08:46 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,962 Thanks: 1606 
\begin{align*} 10(\sin^6{\theta} + \cos^6{\theta}) &= 10(\sin^2{\theta} + \cos^2{\theta})^3  3\sin^4{\theta}\cos^2{\theta}  3\sin^2{\theta}\cos^4{\theta}) \\ &= 10(1  3\sin^2{\theta}\cos^2{\theta}(\sin^2{\theta} + \cos^2{\theta})) \\ &= 10(1  3\sin^2{\theta}\cos^2{\theta}) \end{align*} Thus there were two typos: a second "" instead of "+" in the first line of the original post, and omission of "10" in the second line. 
April 29th, 2018, 05:08 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,308 Thanks: 2443 Math Focus: Mainly analysis and algebra 
And the exponents in the second line make 4.


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$3sin3, cos3theta$, obtained, theta 
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