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Mathematics 22 Online
OpenStudy (anonymous):

Find an exact value. cos(pi/12)

OpenStudy (welshfella):

i think we can use cos 2x = 2cos^2x - 1 cos (pi/6) = 2 cos^ (pi/12) - 1 cos pi/6 = sqrt3/2

OpenStudy (anonymous):

they want you to use the half angle formula \[\cos \frac{ x }{ 2 }= \pm \sqrt{\frac{ 1+cosx }{ 2 }}\]

OpenStudy (welshfella):

oh ok well just plug cos x ( in this case cos pi/6) into the firmula and work it out

OpenStudy (anonymous):

when I do I don't get the right answer

OpenStudy (anonymous):

The answer options are (sqrt6+sqrt2)/4 (-sqrt6+sqrt2)/4 (sqrt6-sqrt2)/4 (-sqrt6-sqrt2)/4

OpenStudy (welshfella):

yes its an awkward one to work out

OpenStudy (anonymous):

How would you work it out?

OpenStudy (welshfella):

= sqrt [( 1 + sqrt3/2)/ 2]

OpenStudy (anonymous):

How do you get from that to one of the answers? I dont understand how you get a sqrt 6 out of any of that

OpenStudy (welshfella):

i simplified that to (1/2)( sqrt(2 + sqrt3))

OpenStudy (welshfella):

i agree I think those answers are wrong.

OpenStudy (welshfella):

because if you go on the calculator cos pi/12 = 0.9659 and my answers comes to the same value

OpenStudy (welshfella):

oh hold on (sqrt6+sqrt2)/4 comes to 0.9659 as well so that one is right

OpenStudy (welshfella):

- also the negative of that is another solution

OpenStudy (welshfella):

I'm not sure how they got to that ...

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