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  • 1 Post By aurel5
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February 3rd, 2015, 01:28 AM   #1
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Wolfram Alpha says:

$ \frac{4\tan 4x\sec 4x}{2\sqrt{\sec 4x}}=2\sin 4x{{\sec }^{\frac{3}{2}}}4x$

I obviously see that 4/2 = 2 but after that I'm just not sure. Can anyone help me out?

Last edited by 3uler; February 3rd, 2015 at 01:30 AM.
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February 3rd, 2015, 03:08 AM   #2
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Originally Posted by 3uler View Post

$ \frac{4\tan 4x\sec 4x}{2\sqrt{\sec 4x}}=2\sin 4x{{\sec }^{\frac{3}{2}}}4x$

$\displaystyle \frac{4\tan 4x\sec 4x}{2\sqrt{\sec 4x}} = \dfrac{4}{2}\cdot\ tan 4x \cdot\dfrac{\sec 4x}{\sqrt{\sec 4x}} = 2 \dfrac{\sin 4x}{\cos 4x}\cdot\dfrac{(\sqrt{\sec 4x})(\sqrt{\sec 4x})}{\sqrt{\sec 4x}}=
=2\sin 4x\cdot\dfrac{1}{\cos4x}\sqrt{\sec 4x} = 2\sin 4x\cdot\sec 4x\cdot\sqrt{\sec 4x}=2\sin 4x\sec 4x(\sec 4x)^{\frac{1}{2}} =2\sin 4x (\sec 4x)^{\frac{3}{2}}$
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