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carmenala2
17.11.2020 •
Mathematics
Consider w=sqrrt2/2(cos(225°) + isin(225°)) and z = 1(cos(60°) + isin(60°)). What is w+ z expressed in rectangular form?
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Ответ:
Option (3)
Step-by-step explanation:
w =![\frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]](/tpl/images/0906/6229/c1990.png)
Since, cos(225) = cos(180 + 45)
= -cos(45) [Since, cos(180 + θ) = -cosθ]
= -![\frac{\sqrt{2}}{2}](/tpl/images/0906/6229/43a11.png)
sin(225) = sin(180 + 45)
= -sin(45)
= -![\frac{\sqrt{2}}{2}](/tpl/images/0906/6229/43a11.png)
Therefore, w =![\frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]](/tpl/images/0906/6229/37205.png)
=![-\frac{2}{4}(1+i)](/tpl/images/0906/6229/77530.png)
=![-\frac{1}{2}(1+i)](/tpl/images/0906/6229/f9aff.png)
z = 1[cos(60) + i(sin(60)]
=![[\frac{1}{2}+i(\frac{\sqrt{3}}{2})](/tpl/images/0906/6229/e1121.png)
=![\frac{1}{2}(1+i\sqrt{3})](/tpl/images/0906/6229/3c4b2.png)
Now (w + z) =![-\frac{1}{2}(1+i)+\frac{1}{2}(1+i\sqrt{3})](/tpl/images/0906/6229/5e4f7.png)
=![-\frac{1}{2}-\frac{i}{2}+\frac{1}{2}+i\frac{\sqrt{3}}{2}](/tpl/images/0906/6229/81bcf.png)
=![\frac{(i\sqrt{3}-i)}{2}](/tpl/images/0906/6229/30573.png)
=![\frac{(\sqrt{3}-1)i}{2}](/tpl/images/0906/6229/0e035.png)
Therefore, Option (3) will be the correct option.
Ответ:
X=14 because it is half of 28
Hope this helps