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Mathematics 26 Online
OpenStudy (anonymous):

Use the following to find sin 2x, cos 2x, and tan 2x. Given: cos x= 5/13, for 3 pi/2

OpenStudy (anonymous):

\[\sin(2x)=2\sin(x)\cos(x)\]

OpenStudy (anonymous):

you know \(\cos(x)=\frac{5}{13}\) but you need \(\sin(x)\) do you know how to find it?

OpenStudy (anonymous):

Yes I do

OpenStudy (anonymous):

|dw:1392779399447:dw|

OpenStudy (anonymous):

ok good, what is \(\sin(x)\) ?

OpenStudy (anonymous):

12/13

OpenStudy (anonymous):

careful

OpenStudy (anonymous):

you are told at the beginning that \(x\) is in quadrant 4, so \[\sin(x)=-\frac{12}{13}\]

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

that was the \[\frac{3\pi}{2}<x<2\pi\] business at the beginning

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

then plug in the numbers and compute \[\sin(2x)=2\sin(x)\cos(x)=2\times \left(-\frac{12}{13}\right)\times \left(\frac{5}{13}\right)\]

OpenStudy (anonymous):

\[\cos(2x)=1-2\cos^2(x)=1-2\times\left(\frac{5}{13}\right)^2\] etc

OpenStudy (anonymous):

clear more or less?

OpenStudy (anonymous):

More

OpenStudy (anonymous):

lol good!

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