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Mathematics 16 Online
OpenStudy (anonymous):

True or false? How do you figure that out? (picture below)

OpenStudy (anonymous):

OpenStudy (shubhamsrg):

what did you do ?

OpenStudy (anonymous):

I haven't done anything, that's the question..

OpenStudy (anonymous):

i think you are supposed to replace \(x\) by \(60\) and see if you get the same answer on both sides

OpenStudy (shubhamsrg):

lol i meant,,what have you tried in the question! :D

OpenStudy (anonymous):

since \(\frac{60}{2}=30\) the left hand side will be \[\tan(30)=\frac{1}{\sqrt 3}\]

OpenStudy (anonymous):

oh, @shubhamsrg I haven't tried anything because I didn't know what I was doing :P @satellite73 I'll try that & see what I get. But I didn't know if I'd have to use the equation tan(x/2)=sqrt((1-cosx)/(1+cosx))

OpenStudy (anonymous):

that is not the way i read the question is does say \(x=60\) right? also i am confused because the denominator is written as \(\sin^2(x)\sin(x)\) which is usually written as \(\sin^3(x)\) is there a typo there?

OpenStudy (anonymous):

Maybe? I just took a screen shot of the question.. but it has said things like that before in other lessons, so they may have meant it that way. & Thanks for helping y'all!

OpenStudy (anonymous):

then i would go with what it says.

OpenStudy (anonymous):

Okay, thanks so much:)

OpenStudy (anonymous):

& I got 4sqrt(3)/3 so I think it's false

OpenStudy (anonymous):

i get http://www.wolframalpha.com/input/?i=%281-cos%2860%29%29%2F%28sin^3%2860%29%29 and it is indeed false

OpenStudy (anonymous):

That's what I meant :P

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