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Ivan
3 years ago
7

Let AB = 24, AC = 10 and BC = 26.

Mathematics
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

a) 24

b) 10

c) 12/13

d) 5/13

e) 12/5

Step-by-step explanation:

a) We can see that the leg opposite <C is AB, and we are given AB = 24

b) We can see the leg adjacent to <C is AC, and we are given that AC = 10

c) The trig function sine is equal to

\frac{opposite}{hypotenuse}

The opposite, AB, is 24, and the hypotenuse, BC, is 26. We can plug those numbers in:

sin(c) = \frac{24}{26} = \frac{12}{13}

d)The trig function cosine is equal to

\frac{adjacent}{hypotenuse}

The adjacent, AC, is 10, and the hypotenuse, BC, is 26. We can plug those numbers in:

cos(c) = \frac{10}{26} = \frac{5}{13}

d)The trig function tangent is equal to

\frac{opposite}{adjacent}

The opposite, AB, is 24, and the adjacent, AC, is 10. We can plug those numbers in:

tan(c) = \frac{24}{10} = \frac{12}{5}

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.
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The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2).

Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

Then,

\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

Now, recall the definition of the complex exponential:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

So,

e^{2i-3} = e^{-3}(\cos 2+i\sin 2)

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2.

Now, substituting this equality in (I) we get

\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2.

Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

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3 years ago
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