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Find the exact value of sin(T -pi/6) when sin(t)=1/3 and cos(t)=Rad8/3

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[sin(a+b)=sin(a)cos(b)+sin(b)cos(a)\]
in this case: a = T and b = -pi/6 but not sure if that fits too well
that might work, even with the funny looking decimal

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Other answers:

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

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