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3 years ago

Find the exact value of each expression, if it defined. ( if answer is undefined, enter undefined) tan (-1)

Mathematics

1 answer:

Triss [41]3 years ago

6 0

Answer:

$tan(-1) \approx -0.02$

Step-by-step explanation:

The given expression is

$tan(-1)$

The tangent of -1 is defined, it's around -0.02.

The tangent is a trigonometric function with a period of $\pi$ , where each period is separated by a vertical asymptote which indicates that the function is not determined through all its domain, that's what the question refers to when it says "if is undefined, enter undefined".

However, at $x=-1$ , the tangent is determined, that means, there's no asymptote on that coordinate, that's why it has a "determined value", which is -0.02 approximately.

$tan(-1) \approx -0.02$

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Any complex number has 2 square roots. Using the polar form from part (d), we have

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Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

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$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

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Read 2 more answers

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timurjin [86]

Answer:

The range for the average low temperatures for the states listed in the chart is B) 24.

Step-by-step explanation:

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56, 55, 65, 64, 55, 44, 54, 55, 68

First we put them in order from least to greatest

44, 54, 55, 55, 55, 56, 64, 65, 68

Range is maximum - minimum

68 is maximum and 44 is minimum

68 - 44 =

Hope this helps you! (:

-Hamilton1757

6 0

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