Question

Perform the indicated operation:10{cos(12)+i sin(12)]*14[cos \left(88)+i sin (88)]Give your answer in polar form:

Answer 1

Answer:

$140\lbrack\cos 100^0+i\sin 100^0\rbrack$

Explanation:

Given the operations

$10\mleft[\cos \mleft(12^{\circ}\mright)+i\sin \mleft(12^{\circ}\mright)\mright]^*14\mleft[\cos \mleft(88^{\circ}\mright)+i\sin \mleft(88^{\circ}\mright)\mright]$

First, we distribute each of the number outside the brackets to obtain:

$=\lbrack10\cos (12^{\circ})+10i\sin (12^{\circ})\rbrack\lbrack14\cos (88^{\circ})+14i\sin (88^{\circ})\rbrack$

Next, we expand:

$=10\cos (12^{\circ})\lbrack14\cos (88^{\circ})+14i\sin (88^{\circ})\rbrack+10i\sin (12^{\circ})\lbrack14\cos (88^{\circ})+14i\sin (88^{\circ})\rbrack$

We open the brackets:

$=140\cos (12^{\circ})\cos (88^{\circ})+140i\sin (88^{\circ})\cos (12^{\circ})+140i\sin (12^{\circ})\cos (88^{\circ})+140i^2\sin (12^{\circ})\sin (88^{\circ})$

Finally, we collect like terms and simplify:

$\begin{gathered} =140\cos (12^{\circ})\cos (88^{\circ})+140i\sin (88^{\circ})\cos (12^{\circ})+140i\sin (12^{\circ})\cos (88^{\circ})-140^{}\sin (12^{\circ})\sin (88^{\circ}) \\ =140\cos (12^{\circ})\cos (88^{\circ})-140^{}\sin (12^{\circ})\sin (88^{\circ})+140i\sin (88^{\circ})\cos (12^{\circ})+140i\sin (12^{\circ})\cos (88^{\circ}) \\ =140\cos (12^0+88^0)+140i(\sin (88^0+12^0) \\ =140\lbrack\cos 100^0+i\sin 100^0\rbrack \end{gathered}$

Note the following:

We applied the trigonometric identities below:

$\begin{gathered} \cos (A+B)=\cos A\cos B-\sin A\sin B \\ \sin (A+B)=\sin A\cos B+\cos A\sin B \\ i^2=-1 \end{gathered}$