Answer:
1) cos (140°-(50°+30°)) = 1/2
2) [tex]sin255^{\circ}=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }[/tex]
3) [tex]cos150^{\circ}=-\frac{ \sqrt{3} }{2}[/tex]
4) , [tex]tan 240^{\circ}=\sqrt{3}[/tex]
Step-by-step explanation:
1) cos (140°-(50°+30°))
Solving:
cos (140°-(50°+30°))
=cos(140°-80°)
=cos(60°)
=1/2
So, cos (140°-(50°+30°)) = 1/2
2) sin 255°
We can write sin 255° as sin(180°+75°)
Solving sing formula sin(a+b)=sin(a)cos(y)+cos(x)sin(y)
[tex]sin(180^{\circ}+75^{\circ})\\=sin(180^{\circ})cos(75^{\circ})+cos(180^{\circ})sin(75^{\circ})\\=0(cos(75^{\circ}))+(-1)(\frac{\sqrt{2}+\sqrt{6} }{\sqrt{4} } )\\=0+(-1)(\frac{\sqrt{2}+\sqrt{6} }{\sqrt{4} } )\\\=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }[/tex]
So, [tex]sin255^{\circ}=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }[/tex]
3) cos 150°
We can solve this using the identity: cos(x)=sin(90°-x)
Solving:
cos 150°= sin(90°-150°)
=sin(-60°)
We know sin(-x)=-sin(x)
=-sin(60°)
=[tex]-\frac{ \sqrt{3} }{2}[/tex]
So, we get [tex]cos150^{\circ}=-\frac{ \sqrt{3} }{2}[/tex]
4) tan 240°
We know tan(180°+x)=tan(x)
tan 240° can be written as tan(180°+60°)
Using the above identity:
tan(180°+60°)=tan(60°) = [tex]\sqrt{3}[/tex]
So, [tex]tan 240^{\circ}=\sqrt{3}[/tex]
