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Mathematics 16 Online
OpenStudy (anonymous):

Is sin 15 degrees is the same as pi/12? I'm suppose to find the exact value of the expression. multiple choice hw

OpenStudy (amistre64):

pi/3 = 60 pi/6 = 30 pi/12 = 15

OpenStudy (amistre64):

180/3 = 60 180/6 = 30 180/12 = 15

OpenStudy (anonymous):

so I got that part right. thanks! ......please continue explanation

OpenStudy (anonymous):

amistre amistre amistre

OpenStudy (amistre64):

sin(15) = sin(45-30)

OpenStudy (anonymous):

\[\frac{\pi}{3}\] is not more 60 than 0 is 32

OpenStudy (amistre64):

sin(a-b) = sinacosb-sinbcosa

OpenStudy (anonymous):

0 degrees Celsius is the same temperature as 32 degrees F, but they are by no means the same number!

OpenStudy (amistre64):

pi 180 ---*---- = 60 3 pi

OpenStudy (anonymous):

true but that hardly means that \[\pi = 180\] \[\pi\] is a number close to 3, and 180 is many times bigger

OpenStudy (amistre64):

sin(15) = sin(45)cos(30) - sin(30)cos(45)

OpenStudy (amistre64):

when converting between degrees and radians; pi = 180

OpenStudy (amistre64):

just as 2pi = 360

OpenStudy (anonymous):

the point is that an angle measured is 15 degrees is the same as an angle measured in \[\frac{\pi}{12}\] radians

OpenStudy (anonymous):

i will be quiet now

OpenStudy (anonymous):

mmm. i think i understand. is the exact value squareroot2 ( squareroot 3 - 1) over 4?

OpenStudy (anonymous):

thank you amistre and satellite! ..just please correct my answer if i'm wrong

OpenStudy (amistre64):

sin(15) = \(\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{\sqrt{2}}\)

OpenStudy (amistre64):

sin(15) = \(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\)

OpenStudy (amistre64):

over 4 is right

OpenStudy (amistre64):

\[\sin(15)=\frac{\sqrt{6}-\sqrt{2}}{4}\] IF I DID IT RIGHT

OpenStudy (amistre64):

caps stuck lol

OpenStudy (anonymous):

nice latex too!

OpenStudy (amistre64):

i think i missed it somewhere

OpenStudy (amistre64):

\[sin(15) = sin(45)cos(30) - sin(30)cos(45)\] \[{\sqrt{2} \over 2}.{\sqrt{3}\over 2}-{1 \over 2}{\sqrt{2}\over 2}\] \[{\sqrt{6}\over 4}-{\sqrt{2}\over4}\]

OpenStudy (anonymous):

thank u i'm doing it on paper..ur much faster

OpenStudy (amistre64):

\[\sqrt{6}-\sqrt{2} \over 4\]

OpenStudy (amistre64):

this an take a few forms; all that equal the same thing but look different based upon how you went thru the stuff

OpenStudy (amistre64):

http://www.wolframalpha.com/input/?i=sin%2815%29

OpenStudy (amistre64):

im right lol

OpenStudy (anonymous):

WOW THANKS!

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