Is sin 15 degrees is the same as pi/12? I'm suppose to find the exact value of the expression. multiple choice hw
pi/3 = 60 pi/6 = 30 pi/12 = 15
180/3 = 60 180/6 = 30 180/12 = 15
so I got that part right. thanks! ......please continue explanation
amistre amistre amistre
sin(15) = sin(45-30)
\[\frac{\pi}{3}\] is not more 60 than 0 is 32
sin(a-b) = sinacosb-sinbcosa
0 degrees Celsius is the same temperature as 32 degrees F, but they are by no means the same number!
pi 180 ---*---- = 60 3 pi
true but that hardly means that \[\pi = 180\] \[\pi\] is a number close to 3, and 180 is many times bigger
sin(15) = sin(45)cos(30) - sin(30)cos(45)
when converting between degrees and radians; pi = 180
just as 2pi = 360
the point is that an angle measured is 15 degrees is the same as an angle measured in \[\frac{\pi}{12}\] radians
i will be quiet now
mmm. i think i understand. is the exact value squareroot2 ( squareroot 3 - 1) over 4?
thank you amistre and satellite! ..just please correct my answer if i'm wrong
sin(15) = \(\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{\sqrt{2}}\)
sin(15) = \(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\)
over 4 is right
\[\sin(15)=\frac{\sqrt{6}-\sqrt{2}}{4}\] IF I DID IT RIGHT
caps stuck lol
nice latex too!
i think i missed it somewhere
\[sin(15) = sin(45)cos(30) - sin(30)cos(45)\] \[{\sqrt{2} \over 2}.{\sqrt{3}\over 2}-{1 \over 2}{\sqrt{2}\over 2}\] \[{\sqrt{6}\over 4}-{\sqrt{2}\over4}\]
thank u i'm doing it on paper..ur much faster
\[\sqrt{6}-\sqrt{2} \over 4\]
this an take a few forms; all that equal the same thing but look different based upon how you went thru the stuff
im right lol
WOW THANKS!
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