Question

What is the quotient of startstartfraction 90 (cosine (startfraction pi over 4 endfraction) i sine (startfraction pi over 4 endfraction) ) overover 2 (cosine (startfraction pi over 12 endfraction) i sine (startfraction pi over 12 endfraction) ) endendfraction ?

Answer 1

The value of the quotient is  $\frac{45\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})} = 45(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$

How to determine the quotient?

The expression is given as:

$\frac{90\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{2\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})}$

Divide 90 by 2

$\frac{45\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})}$

As a general rule, we have:

$\frac{\cos(\frac{\pi}{A}) i\sin(\frac{\pi}{A})}{\cos(\frac{\pi}{B}) i\sin(\frac{\pi}{B})} = \cos(\frac{\pi}{2B/A}) + i\sin(\frac{\pi}{2B/A})$

The above means that:

A = 4 and B = 12

So, we have:

2B/A = 2 * 12/4

Evaluate

2B/A = 6

So, the equation becomes

$\frac{\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})} = \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})$

Substitute $\frac{\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})} = \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})$ in $\frac{45\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})}$

$\frac{45\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})} = 45(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$

Hence, the value of the quotient is  $\frac{45\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})} = 45(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$

Read more about trigonometry expressions at:

brainly.com/question/561827

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Complete question

What is the quotient of $\frac{90\cos(\frac{\pi}{4}) i\sin(\frac{\pi}{4})}{2\cos(\frac{\pi}{12}) i\sin(\frac{\pi}{12})}$