Finding the exact value

[MathsHelpPlz]

"Find the exact value of (4sin149 + 12sin150 + 4sin151)/(3cos149 + 9cos150 + 3cos151)"

I took out the common factor of 4 from top and 3 from bottom but I don't know what to do next.

Thanks for your time.

EDIT: Actually sorry I understand it, the question had previously told us that a similar expression equals the tangent of the middle number (150 in this case) so its 4/3 tan150

 

[MarkFL]

\(\displaystyle \dfrac{4 \sin(149^{ \circ})+12 \sin(150^{ \circ})+4 \sin(151^{ \circ})}{3 \cos(149^{ \circ})+9 \cos(150^{ \circ})+3 \cos(151^{ \circ})}=\)

\(\displaystyle \dfrac{4\left( \sin(149^{ \circ})+ \frac{3}{2}+ \sin(151^{ \circ}) \right)}{3\left( \cos(149^{ \circ})-\frac{3 \sqrt{3}}{2}+ \cos(151^{ \circ}) \right)}=\)

\(\displaystyle \dfrac{4\left( \cos(1^{ \circ})+ \frac{3}{2} \right)}{3\left(- \sqrt{3} \cos(1^{ \circ})- \frac{3 \sqrt{3}}{2} \right)}=\)

\(\displaystyle \dfrac{4(2 \cos(1^{ \circ})+3)}{-3 \sqrt{3}(2 \cos(1^{ \circ})+3)}=\)

\(\displaystyle -\dfrac{4}{3 \sqrt{3}}\)

 

[HallsofIvy]

In case you were wondering, MarkFL is using the fact that 150= 90+ 60 as well as the identities cos(a+b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b). So cos(150)= cos(90+ 60)= cos(90)cos(60)- sin(90)sin(60) and sin(150)= sin(90+ 60)= sin(90)cos(60)+ cos(90)sin(60). And you certainly should know that know that sin(90)= 1 while cos(90)= 0 so that cos(150)= 0(cos(60))- 1(sin(60))= -sin(60) and sin(150)= 1(cos(60))+ 0(sin(60))= cos(60).

\(\displaystyle sin(60)= \sqrt{3}/2\) and \(\displaystyle cos(60)= 1/2\) are also values you should know or be able to get (bisect an equilateral triangle to get two 30-60-90 right triangles). That is, \(\displaystyle cos(150)= -\sqrt{3}/2\) and \(\displaystyle sin(150)= 1/2\).