anonymous
  • anonymous
Find the exact value of sin(T -pi/6) when sin(t)=1/3 and cos(t)=Rad8/3
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
\[sin(a+b)=sin(a)cos(b)+sin(b)cos(a)\]
amistre64
  • amistre64
in this case: a = T and b = -pi/6 but not sure if that fits too well
amistre64
  • amistre64
that might work, even with the funny looking decimal

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anonymous
  • anonymous
ops there is a error there i will correct that
myininaya
  • myininaya
what funny looking decimal?
myininaya
  • myininaya
T=t?
anonymous
  • anonymous
\[\sin(T - \pi/6) = \sin(T)\cos(\pi/6)-\sin(\pi/6)\cos(T)\]\[\sqrt{3}/6 - 0,0232 = 0,2654\]
anonymous
  • anonymous
so basically use the double angle formula?
anonymous
  • anonymous
yes
myininaya
  • myininaya
\[\frac{1}{3} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{\sqrt{8}}{3}\]
anonymous
  • anonymous
oh did you get that from a chart myininaya?
anonymous
  • anonymous
Remember that \[\sin(a \pm b)= \sin(a)\cos(b) \pm \sin(b)\cos(a)\]
anonymous
  • anonymous
ok.. yes
myininaya
  • myininaya
yep bella \[\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}; \sin(\frac{\pi}{6})=\frac{1}{2}\]
anonymous
  • anonymous
ok thank you!
amistre64
  • amistre64
across the pond they use the "," as a "."

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