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Mathematics 15 Online
OpenStudy (anonymous):

how does (cos pi/12 X cos pi/12) - (sin pi/12 x sin pi/12) = pi/6?

OpenStudy (freckles):

\[\cos(\frac{\pi}{12}) \cdot \cos(\frac{\pi}{12}) -\sin(\frac{\pi}{12}) \cdot \sin(\frac{\pi}{12})=\cos^2(\frac{\pi}{12})-\sin^2(\frac{\pi}{12})\] Do you recall the following: \[\cos^2(x)-\sin^2(x)=\cos(2x)\]

OpenStudy (freckles):

So here we clearly have \[x=\frac{\pi}{12} => 2x=?\]

OpenStudy (freckles):

@shilohk ?

OpenStudy (freckles):

Do you understand? Do you know what 2x is if x=pi/12?

OpenStudy (mertsj):

It doesn't equal pi/6. It equals cos(pi/6)

OpenStudy (anonymous):

Freckles, yes, it would be 2pi/12 or pi/6... thank you!

OpenStudy (freckles):

:) Awesome shilohk! :)

OpenStudy (freckles):

So we have cos(pi/6)

OpenStudy (freckles):

So do you know how to evaluate that?

OpenStudy (freckles):

@Mertsj I was working with the inside first

OpenStudy (freckles):

The inside is pi/6 since x is pi/12 and 2x is 2pi/12=pi/6

OpenStudy (mertsj):

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