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Divide.

32(cos(7π/4)+i sin(7π/4))/4√2(cos(5π/6)+i sin(5π/6))

___(cos(___)+i sin(___))​

Divide. 32(cos(7π/4)+i sin(7π/4))/4√2(cos(5π/6)+i sin(5π/6)) ___(cos(___)+i sin(___))​-example-1

1 Answer

6 votes

Answer:

-√6(cos(π+π/6)+i sin(π+π/6))

or

-2.449(cos(3.658)+i sim( 3.658)

Explanation:

First, let's simplify the expression in the denominator:

4√2(cos(5π/6)+i sin(5π/6)) = 4√2(√3/2 + i/2) = 2√6 + 2i√2

Now we can rewrite the entire expression as:

32(cos(7π/4)+i sin(7π/4)) / (2√6 + 2i√2)

To simplify the division, we can multiply both the numerator and denominator by the conjugate of the denominator, which is:

2√6 - 2i√2

Multiplying the numerator and denominator by this conjugate, we get:

32(cos(7π/4)+i sin(7π/4)) * (2√6 - 2i√2) / [(2√6 + 2i√2) * (2√6 - 2i√2)]

Simplifying the denominator on the right side gives us:

(2√6)^2 - (2i√2)^2 = 24 - 8 = 16

So we can simplify the expression further to:

32(cos(7π/4)+i sin(7π/4)) * (2√6 - 2i√2) / 16

Canceling out the 16 in the denominator and simplifying the numerator gives:

2(cos(7π/4)+i sin(7π/4)) * (√6 - i√2)

Now we can expand the complex multiplication in the numerator:

2(cos(7π/4)√6 - sin(7π/4)√2 + i(cos(7π/4)√2 + sin(7π/4)√6))

Simplifying the trigonometric functions using the values of sine and cosine for 7π/4 (which is equivalent to -π/4), we get:

2(-√6/2 - √2/2i + i√2/2 - √6/2i)

Simplifying further, we can combine like terms and rationalize the denominator:

-√6 - √3i

Therefore, the final answer is:

-√6(cos(π+π/6)+i sin(π+π/6))

Hope this helps!

User Jashaszun
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