sin 2x + cos60 = 0 ?
i have found sin inverse -1/2 as -30, how do i get it positive in the third and fourth quadrant?
hold on you have \[\sin(2x)=\frac{1}{2}=0\] rigth?
so \[\sin(2x)=-\frac{1}{2}\]
you are working in degrees, so put away the calculator for a moment, we can use it later
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we want the angle whose sine is -1/2 and you can see that there are two of them. your calculator gives you -30, but that is the same as 330 degrees
however if you only use your calculator you are going to miss the one at 210 degrees
so you cannot only rely on your calculator for these. to finish you have \[2x=210\] \[x=105\] or \[2x=330\] \[x=165\]
how is it 210? won't it be 270 - 30?
i was counting as 180 + 30
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oh, but how? i thought that is used for tan only, and for sin it's 180 + (-30)?? it's all a bit confusing for me :P
lets go slow
forgetting about the 2x for a moment, you are looking for an angle measured in degrees, whose sine is -1/2
that means the second coordinate on the unit circle is -1/2
there are clearly two of them, one in quadrant 3 and one in quadrant 4. the one in quadrant 4 is -30 degrees and also 330 degrees. they are the same place on the unit cirlce
then there is the one directly across as in the rather lame picture i drew. directly across the circle. now i am not trying to use a formula or any property of sine or whatever to find that angle. i am just thinking "what is it?" or counting out in multiples of 30 degrees.30, 60, 90, 120 , 150 , 180 , 210 until i get there
it is better to have a picture and write down the point you are trying to find than to try to do it in the abstract. then you won't be confused by "adding 180" or finding complimentary or supplementary angles or whatever else they have you doing.
naww, it's not lame, it's cool. and thanks, i think i got it :)
good!
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