Answer:
Part A:
The cosine sum identity is given below as
[tex]\cos(A+B)=\cos A\cos B-\sin A\sin B[/tex]
For cos 240, we will have
[tex]\begin{gathered} \cos240=cos(180+60) \\ \end{gathered}[/tex]
By applying the rule, we will have
[tex]\begin{gathered} \cos(180+60)=\cos180\cos60-\sin180\sin60 \\ \cos(180+60)=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos240=cos(180+60)=-\frac{1}{2} \end{gathered}[/tex]
Hence,
The value of cos 240 is
[tex]\Rightarrow\cos240=cos(180+60)=-\frac{1}{2}[/tex]
Part B:
The sine difference identity is given below as
[tex]\sin(A-B)=\sin A\cos B-\sin B\cos A[/tex][tex]\sin240=sin(360-120)[/tex][tex]\begin{gathered} \sin(A-B)=\sin(A)\cos(B)-\sin(B)\cos(A) \\ sin240=sin(360-120)=sin360cos120-sin120cos360 \\ sin240=sin(360-120)=0(cos120)-1(sin120) \\ sin240=sin(360-120)=-sin120 \\ sin240=sin(360-120)=-sin(180-120) \\ sin240=sin(360-120)=-sin60 \\ sin240=sin(360-120)=-\frac{\sqrt{3}}{2} \end{gathered}[/tex]
Hence,
The final answer of sin240 is
[tex]\Rightarrow-\frac{\sqrt{3}}{2}[/tex]
