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frutty [35]
3 years ago
7

Simplify the expression square root of -1 over (3+8i) - (2+5i)

Mathematics
1 answer:
Lynna [10]3 years ago
8 0

\frac{i}{(3 \:  +  \: 8i) \:  -  \: (2 \:  +  \: 5i)}  \:  =  \:  \frac{i}{1 \:  +  \: 3i}  \:  =  \:  \frac{i}{1 \:  +  \: 3 i}  \:  \times  \:  \frac{1 \:  -  \: 3i}{1 \:  -  \: 3i}  \:  =  \:  \frac{i \:  +  \: 3}{1 \:  +  \: 9}  \:  =  \:  \frac{3 \:  +  \: i}{10}A) (3 + i)/10
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See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

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\cos(x)=  \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}

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======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3

\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0

t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}

Just note that

\tan\left(\dfrac{x}{2}\right) =  \dfrac{3\pm 8\sqrt{10}}{-2}

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Answer: The fabric costs $7.60 per yard.

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