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dezoksy [38]
3 years ago
9

Examine the pattern in the powers of i you wrote in the table, and create a rule for finding the value of large powers of i. Jus

tify your answer.
Mathematics
1 answer:
denpristay [2]3 years ago
5 0
Let N be any integer. Then

i^N=\begin{cases}1&\text{if }N\equiv0\mod4\\i&\text{if }N\equiv1\mod4\\-1&\text{if }N\equiv2\mod4\\-i&\text{if }N\equiv3\mod4\end{cases}

In other words:

(1) If N is a multiple of 4, then i^N=1.

(2) If dividing N by 4 leaves a remainder of 1, then i^N=i, since i^N=i^{4n+1} for some integer n, and from (1) we know that i^{4n+1}=i^{4n}i=i (because, obviously, 4n is a multiple of 4).

(3) If instead you get a remainder of 2, then i^N=-1. This follows from (1) as well. i^N=i^{4n+2}=i^{4n}i^2=(1)(-1)=-1.

(4) Finally, if you get a remainder of 3, then i^N=i^{4n+3}=i^{4n}i^2i=(1)(-1)(i)=-i.
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x is the number of days worked

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A triangle has sides of length 3 cm and 3 cm. What can be concluded about the length of the third side?
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Find the value of x round to the nearest tenth 11.2 12.6 17.5 35.7
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Read 2 more answers
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
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