Question

Hello! Need some help on parts A and B, the rubric and question is attached. Thank you!

Answer 1

Part A.

We need to find cos 240º using the cosine sum identify.

This identity is given by the formula:

`\cos(a+b)=\cos a\cos b-\sin a\sin b`

Now, we can write:

`\cos240\degree=\cos(180\degree+60\degree)`

Then, we obtain:

`\begin{gathered} \cos240\degree=\cos(180\degree+60\degree) \\ \\ \cos240\degree=\cos180\degree\cos60\degree-\sin180\degree\sin60\degree \\ \\ \cos240\degree=-1\cdot(1)/(2)-0\cdot(√(3))/(2) \\ \\ \cos240\degree=-(1)/(2) \\ \\ \cos240\degree=-0.5 \end{gathered}`

Part B.

We need to find sin 240º using the sine difference identity.

This identity is given by:

`\sin(a-b)=\sin a\cos b-\sin b\cos a`

We can write:

`\sin240\degree=\sin(270\degree-30\degree)`

Then, we obtain:

`\begin{gathered} \sin240\degree=\sin270\degree\cos30\degree-\sin30\degree\cos270\degree \\ \\ \sin240\degree=-1\cdot(√(3))/(2)-(1)/(2)\cdot0 \\ \\ \sin240\degree=-(√(3))/(2) \end{gathered}`

Answers

`\begin{gathered} \text{ Part A. }\cos240\degree=-(1)/(2) \\ \\ \text{ Part B. }\sin240\degree=-(√(3))/(2) \end{gathered}`