Question

Can somebody explain how trigonometric form polar equations are divided/multiplied?

Answer 1

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.