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April 28th, 2018, 06:39 PM   #1
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How $3sin^3 \theta cos^3\theta$ obtained?

$15 (\sin^4X + \cos^4X) - 10 (\sin^6X - \cos^6X) = ?$

$\rightarrow 15 (1-2 \sin^2 \theta. \cos^2\theta) - (1-3\sin^3 \theta \cos^3\theta)$

$\rightarrow 15 - 30 \sin^2 \theta. \cos^2\theta - 10 + 30 \sin^2 \theta. \cos^2\theta$

$\rightarrow 15-10 = 5$

How $3\sin^3 \theta \cos^3\theta$ obtained?

Last edited by skipjack; April 28th, 2018 at 08:01 PM.
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April 28th, 2018, 08:16 PM   #2
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Quote:
Originally Posted by Ganesh Ujwal View Post
How $3\sin^3 \theta \cos^3\theta$ obtained?
You are likely running into trouble because it isn't true.

-Dan
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April 28th, 2018, 08:20 PM   #3
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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*}
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April 28th, 2018, 08:46 PM   #4
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\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.
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April 29th, 2018, 05:08 AM   #5
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And the exponents in the second line make 4.
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