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write the expression as the sine, cosine, or tangent of a double angle Then find the exact value of the expression. 2 sin 120 cos 120

 Jul 23, 2014

Best Answer 

 #2
avatar+118608 
+5

$$2sin120cos120\\
=sin(120+120)\\
=sin240\\
=sin(180+60) \qquad (third\;\;quadrant\;\;so\;\;negative)\\
=-sin60\\
=\frac{-\sqrt3}{2}$$

.
 Jul 23, 2014
 #1
avatar+26367 
+5

2 sin 120 cos 120    ?

$$\\\sin{(\alpha+\beta)}=
\sin{(\alpha)}
*
\cos{(\beta)}
+
\cos{(\alpha)}
*
\sin{(\beta)}
\\
\sin{(\alpha+\alpha)}=
\sin{(\alpha)}
*
\cos{(\alpha)}
+
\cos{(\alpha)}
*
\sin{(\alpha)}\\\\
\boxed{
\sin{(2*\alpha)}=
2*\sin{(\alpha)}
*
\cos{(\alpha)}}\\\\
\sin{(2*120\ensurement{^{\circ}} )}=
2\sin{(120\ensurement{^{\circ}} )}
\cos{(120\ensurement{^{\circ}} )}=
\sin{(240 \ensurement{^{\circ}} ) =-0.86602540378$$

$$\\ \sin{(240 \ensurement{^{\circ}} )
=
\sin{(360\ensurement{^{\circ}-120 \ensurement{^{\circ} } )
=
-\sin{(120 \ensurement{^{\circ} } )
=-\frac{\sqrt{3} } {2}
=-0.86602540378$$

 Jul 23, 2014
 #2
avatar+118608 
+5
Best Answer

$$2sin120cos120\\
=sin(120+120)\\
=sin240\\
=sin(180+60) \qquad (third\;\;quadrant\;\;so\;\;negative)\\
=-sin60\\
=\frac{-\sqrt3}{2}$$

Melody Jul 23, 2014

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