Question

i need as much shown work as possible and explanation please

Answer 1

Answer:

Part A:

The cosine sum identity is given below as

$\cos(A+B)=\cos A\cos B-\sin A\sin B$

For cos 240, we will have

$\begin{gathered} \cos240=cos(180+60) \\ \end{gathered}$

By applying the rule, we will have

$\begin{gathered} \cos(180+60)=\cos180\cos60-\sin180\sin60 \\ \cos(180+60)=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos240=cos(180+60)=-\frac{1}{2} \end{gathered}$

Hence,

The value of cos 240 is

$\Rightarrow\cos240=cos(180+60)=-\frac{1}{2}$

Part B:

The sine difference identity is given below as

$\sin(A-B)=\sin A\cos B-\sin B\cos A$$\sin240=sin(360-120)$$\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}$

Hence,

The final answer of sin240 is

$\Rightarrow-\frac{\sqrt{3}}{2}$