I am not sure how to solve this, PLEASE HELP ---pi/3sinpi/6=1/2(sinpi/2-sinpi/6)
\[\large \frac{\pi}{3sin(\frac{\pi}{6})} = \frac{1}{2(sin(\frac{\pi}{2} -sin(\frac{\pi}{6})))}\] Not sure what your question is supposed to look like..lol like this?
nooo, the lines at the begining I think are there to say that something is missing in the equation to make both sides equal. its "blank" pi/3 sin pi/6 = 1/2 (sin pi/2 - sin pi/6)
\["blank" \pi/3 \sin \pi/6 = 1/2(\sin \pi/2 - \sin \pi/6)\]
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Oh! okay so \[\large \text{"Blank"}\frac{\pi}{3}sin\frac{\pi}{6} = \frac{1}{2}(sin(\frac{\pi}{2}) - sin(\frac{\pi}{6})))\] Well, lets just simplify a bit using what we know \(\large sin(\frac{\pi}{6}) = sin(30) = \frac{1}{2}\) \(\large sin(\frac{\pi}{2}) = sin(90) = 1\) So \[\large \text{"Blank"}\frac{\pi}{3}(\frac{1}{2}) = \frac{1}{2}(1 - \frac{1}{2})\] \[\large \text{"Blank"}\frac{\pi}{3}(\frac{1}{2}) = \frac{1}{2}(\frac{1}{2})\] \[\large \text{"Blank"}\frac{\pi}{3}(\frac{1}{2}) = \frac{1}{4}\] \[\large \text{"Blank"}\frac{\pi}{3} = \frac{1}{2}\] So now, what trig function when evaluated at pi/3 gives you 1/2??
Tell me if any lines confuse you!! :)
Oh I see so it is either going to be sin, cos, tan, cot, or sec?
Right! It makes sense right? you want to make the Left hand side = Right hand side...so we definitely need to have a trig function be the answer
alright and by looking at the unit curcle, you can see that it is indeed cosine of pi/3 that is equal to 1/2
Indeed :P So I *just I* would assume the answer they want, is the blank needs to be replaced by cos
can you help with another one?
I actually have to head out to work now hun :/ sorry! But there are many others who I know can help you :D
okay thats fine!
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