204k views
3 votes
Can somebody explain how trigonometric form polar equations are divided/multiplied?

Can somebody explain how trigonometric form polar equations are divided/multiplied-example-1
Can somebody explain how trigonometric form polar equations are divided/multiplied-example-1
Can somebody explain how trigonometric form polar equations are divided/multiplied-example-2

1 Answer

6 votes

Answer:

Attachment 1 : Option C

Attachment 2 : Option A

Explanation:

( 1 ) Expressing the product of z1 and z2 would be as follows,


14\left[\cos \left((\pi \:)/(5)\right)+i\sin \left((\pi \:\:)/(5)\right)\right]\cdot \:2√(2)\left[\cos \left((3\pi \:)/(2)\right)+i\sin \left((3\pi \:\:)/(2)\right)\right]

Now to solve such problems, you will need to know what cos(π / 5) is, sin(π / 5) etc. If you don't know their exact value, I would recommend you use a calculator,

cos(π / 5) =
(√(5)+1)/(4),

sin(π / 5) =
\frac{√(2)\sqrt{5-√(5)}}{4}

cos(3π / 2) = 0,

sin(3π / 2) = - 1

Let's substitute those values in our expression,


14\left[(√(5)+1)/(4)+i\frac{√(2)\sqrt{5-√(5)}}{4}\right]\cdot \:2√(2)\left[0-i\right]

And now simplify the expression,


14\sqrt{5-√(5)}+i\left(-7√(10)-7√(2)\right)

The exact value of
14\sqrt{5-√(5)} =
23.27510\dots and
(-7√(10)-7√(2)\right)) =
-32.03543\dots Therefore we have the expression
23.27510 - 32.03543i, which is close to option c. As you can see they approximated the solution.

( 2 ) Here we will apply the following trivial identities,

cos(π / 3) =
(1)/(2),

sin(π / 3) =
(√(3))/(2),

cos(- π / 6) =
(√(3))/(2),

sin(- π / 6) =
-(1)/(2)

Substitute into the following expression, representing the quotient of the given values of z1 and z2,


15\left[cos\left((\pi \:)/(3)\right)+isin\left((\pi \:\:)/(3)\right)\right] / \:3√(2)\left[cos\left((-\pi \:)/(6)\right)+isin\left((-\pi \:\:)/(6)\right)\right]


15\left[(1)/(2)+(√(3))/(2)\right]/ \:3√(2)\left[(√(3))/(2)+-(1)/(2)\right]

The simplified expression will be the following,


i(5√(2))/(2) or in other words
(5√(2))/(2)i or
(5i√(2))/(2)

The solution will be option a, as you can see.

User Tony Park
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.