Question

Find the exact value of the expression.
sin(7pi/6) * sin(pi/3)

I get (-1 - sqrt3)/2, but these are the answer choices:

A. (1 + sqrt3)/2
B. -((1 + sqrt3)/2)
C. -((2 + sqrt2)/3

Answer 1

your answer matches B

Answer 2

\[2\sin A \sin B=\cos (A-B)-\cos (A+B)\]
\[2\sin \frac{ 7 \pi }{ 6 }\sin \frac{ \pi }{ 3 }=\cos \left( \frac{ 7 \pi }{ 6 }-\frac{ \pi }{ 3 } \right)-\cos \left( \frac{ 7 \pi }{ 6 }+\frac{ \pi }{ 3 } \right)\]
\[=\cos \frac{ 5 \pi }{ 6 }+\cos \left( \frac{ 9 \pi }{ 6 } \right)\]
\[=\cos \left( \pi-\frac{ \pi }{ 6 } \right)-\cos \left( \pi+\frac{ \pi }{ 2 } \right)\]
\[=-\cos \frac{ \pi }{ 6 }-\cos \frac{ \pi }{ 2 }=-\frac{ \sqrt{3} }{ 2 }-0\]
\[\sin \frac{ 7 \pi }{ 6 }\sin \frac{ \pi }{ 3 }=-\frac{ \sqrt{3} }{ 4 }\]

Answer 3

How you get that answer?

Answer 4

Either your statement is wrong or options are wrong.

Answer 5

I'm so sorry...it says sin(7pi/6) - sin(pi/3)

Answer 6

\[\sin \frac{ 7 \pi }{ 6 }-\sin \frac{ \pi }{ 3 }=\sin \left( \pi+\frac{ \pi }{ 6 } \right)-\sin \frac{ \pi }{ 3 }\]
=-\[=-\sin \frac{ \pi }{ 6 }-\sin \frac{ \pi }{ 3 } =-\frac{ 1 }{ 2 }-\frac{ \sqrt{3} }{ 2 }=-\frac{ 1+\sqrt{3} }{ 2 }\]

Answer 7

Yes, I see now

Answer 8

Thank you!

Answer 9

yw