Part A.
We need to find cos 240º using the cosine sum identify.
This identity is given by the formula:
[tex]\cos(a+b)=\cos a\cos b-\sin a\sin b[/tex]
Now, we can write:
[tex]\cos240\degree=\cos(180\degree+60\degree)[/tex]
Then, we obtain:
[tex]\begin{gathered} \cos240\degree=\cos(180\degree+60\degree) \\ \\ \cos240\degree=\cos180\degree\cos60\degree-\sin180\degree\sin60\degree \\ \\ \cos240\degree=-1\cdot\frac{1}{2}-0\cdot\frac{\sqrt{3}}{2} \\ \\ \cos240\degree=-\frac{1}{2} \\ \\ \cos240\degree=-0.5 \end{gathered}[/tex]
Part B.
We need to find sin 240º using the sine difference identity.
This identity is given by:
[tex]\sin(a-b)=\sin a\cos b-\sin b\cos a[/tex]
We can write:
[tex]\sin240\degree=\sin(270\degree-30\degree)[/tex]
Then, we obtain:
[tex]\begin{gathered} \sin240\degree=\sin270\degree\cos30\degree-\sin30\degree\cos270\degree \\ \\ \sin240\degree=-1\cdot\frac{\sqrt{3}}{2}-\frac{1}{2}\cdot0 \\ \\ \sin240\degree=-\frac{\sqrt{3}}{2} \end{gathered}[/tex]
Answers
[tex]\begin{gathered} \text{ Part A. }\cos240\degree=-\frac{1}{2} \\ \\ \text{ Part B. }\sin240\degree=-\frac{\sqrt{3}}{2} \end{gathered}[/tex]
