anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
pi/3 = 60 pi/6 = 30 pi/12 = 15
amistre64
  • amistre64
180/3 = 60 180/6 = 30 180/12 = 15
anonymous
  • anonymous
so I got that part right. thanks! ......please continue explanation

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anonymous
  • anonymous
amistre amistre amistre
amistre64
  • amistre64
sin(15) = sin(45-30)
anonymous
  • anonymous
\[\frac{\pi}{3}\] is not more 60 than 0 is 32
amistre64
  • amistre64
sin(a-b) = sinacosb-sinbcosa
anonymous
  • anonymous
0 degrees Celsius is the same temperature as 32 degrees F, but they are by no means the same number!
amistre64
  • amistre64
pi 180 ---*---- = 60 3 pi
anonymous
  • anonymous
true but that hardly means that \[\pi = 180\] \[\pi\] is a number close to 3, and 180 is many times bigger
amistre64
  • amistre64
sin(15) = sin(45)cos(30) - sin(30)cos(45)
amistre64
  • amistre64
when converting between degrees and radians; pi = 180
amistre64
  • amistre64
just as 2pi = 360
anonymous
  • anonymous
the point is that an angle measured is 15 degrees is the same as an angle measured in \[\frac{\pi}{12}\] radians
anonymous
  • anonymous
i will be quiet now
anonymous
  • anonymous
mmm. i think i understand. is the exact value squareroot2 ( squareroot 3 - 1) over 4?
anonymous
  • anonymous
thank you amistre and satellite! ..just please correct my answer if i'm wrong
amistre64
  • amistre64
sin(15) = \(\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{\sqrt{2}}\)
amistre64
  • amistre64
sin(15) = \(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\)
amistre64
  • amistre64
over 4 is right
amistre64
  • amistre64
\[\sin(15)=\frac{\sqrt{6}-\sqrt{2}}{4}\] IF I DID IT RIGHT
amistre64
  • amistre64
caps stuck lol
anonymous
  • anonymous
nice latex too!
amistre64
  • amistre64
i think i missed it somewhere
amistre64
  • amistre64
\[sin(15) = sin(45)cos(30) - sin(30)cos(45)\] \[{\sqrt{2} \over 2}.{\sqrt{3}\over 2}-{1 \over 2}{\sqrt{2}\over 2}\] \[{\sqrt{6}\over 4}-{\sqrt{2}\over4}\]
anonymous
  • anonymous
thank u i'm doing it on paper..ur much faster
amistre64
  • amistre64
\[\sqrt{6}-\sqrt{2} \over 4\]
amistre64
  • amistre64
this an take a few forms; all that equal the same thing but look different based upon how you went thru the stuff
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sin%2815%29
amistre64
  • amistre64
im right lol
anonymous
  • anonymous
WOW THANKS!

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