How do you simplify #2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))# and express the result in rectangular form?

Question

Sadie Ackerman · Accepted Answer

#2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))=-2-2i# #2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))# = #2sqrt2{cos((3pi)/4)cos(pi/2)+icos((3pi)/4)sin(pi/2)+icos(pi/2)sin((3pi)/4)+i^2sin((3pi)/4)sin(pi/2)}# = #2sqrt2{cos((3pi)/4)cos(pi/2)-sin((3pi)/4)sin(pi/2)+i(cos((3pi)/4)sin(pi/2)+icos(pi/2)sin((3pi)/4))}# = #2sqrt2{cos((3pi)/4+pi/2)+isin((3pi)/4+pi/2)}# = #2sqrt2(cos((5pi)/4)+isin((5pi)/4))# = #2sqrt2(-cos(pi/4)-isin(pi/4))# = #2sqrt2(-1/sqrt2-i1/sqrt2)# = #-2-2i#

Sophie Adams · Answer

undefined To simplify the expression and express the result in rectangular form, you first multiply the terms inside the parentheses, then use Euler's formula $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

1. Multiply the terms inside the parentheses:

$$2(\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})) \times \sqrt{2}(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}))$$

$$= 2\sqrt{2}(\cos(\frac{3\pi}{4})\cos(\frac{\pi}{2}) - \sin(\frac{3\pi}{4})\sin(\frac{\pi}{2}) + i(\cos(\frac{3\pi}{4})\sin(\frac{\pi}{2}) + \sin(\frac{3\pi}{4})\cos(\frac{\pi}{2})))$$

2. Use angle addition identities for cosine and sine:

$$= 2\sqrt{2}(\cos(\frac{3\pi}{4} + \frac{\pi}{2}) + i\sin(\frac{3\pi}{4} + \frac{\pi}{2}))$$

3. Simplify the angles:

$$= 2\sqrt{2}(\cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4}))$$

4. Convert back to rectangular form:

$$= 2\sqrt{2}(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i)$$

$$= 2(\sqrt{2} - i\sqrt{2})$$

$$= 2\sqrt{2} - 2i\sqrt{2}$$

So, the result expressed in rectangular form is $2\sqrt{2} - 2i\sqrt{2}$.