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March 8th, 2018, 03:55 PM   #1
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May someone please help me with this question?

Hello, I would like to know if you can help me with this problem the intergal of (cos (6x)+1), I’ve understood it as far as cos (1/2 (ax)) but where is the sqrt of 2 coming from?! Thank you.
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March 8th, 2018, 05:03 PM   #2
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\begin{align*}\cos{(2A)} &= \cos^2{(A)} - \sin^2{(A)} \\
&= \cos^2{(A)} - \big(1 - \cos^2{(A)}\big) \\
&= 2\cos^2{(A)} - 1 \\
\cos{(2A)} + 1 &= 2\cos^2{(A)} \\[8pt]
\sqrt{\cos{(2A)} + 1} &= \sqrt{2\cos^2{(A)}} \\
&= \sqrt{2\vphantom{\cos^2{(A)}}}\sqrt{\cos^2{(A)}} \\
&= \sqrt{2}\cos{(A)}\end{align*}

Last edited by v8archie; March 8th, 2018 at 05:05 PM.
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March 8th, 2018, 05:17 PM   #3
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Oh my goodness, thank you so much for your help, I greatly appreciate it. I’ve been trying to figure this out for so long. I have and solution, but is this the same as this square root of 2, I understand where the 1/3 (4 square root of 2 comes from).
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March 8th, 2018, 06:53 PM   #4
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The $\sqrt2$ comes from the derivation above. It gets multiplied by the $1$ in the following.
\begin{align*}\sin{(3x)} &= \sin{(2x)}\cos{(x)} + \cos{(2x)}\sin{(x)} \\
&= \big(2\sin{(x)}\cos{(x)}\big)\cos{(x)} + \big(2\cos^2{(x)}-1\big)\sin{(x)} \\
&= 4\sin{(x)}\cos^2{(x)} - \sin{(x)} \\
&= \big(4\cos^2{(x)} - 1\big)\sin{(x)} \end{align*}
I don't know why Symbolab decided to write the answer in trigonometric functions of $x$. I'd have left it in terms of trigonometric functions of $3x$.

Last edited by v8archie; March 8th, 2018 at 06:55 PM.
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