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Answer:

#(1+7i)/(5-3i)=5/sqrt17(costheta+isintheta)#, where #theta=tan^(-1)(-19/8)#

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

#z_1=r_1(cosalpha+isinalpha)# and #z_2=r_2(cosbeta+isinbeta)#

Here, if two complex numbers are #a_1+ib_1# and #a_2+ib_2# #r_1=sqrt(a_1^2+b_1^2)#, #r_2=sqrt(a_2^2+b_2^2)# and #alpha=tan^(-1)(b_1/a_1)#, #beta=tan^(-1)(b_2/a_2)#

Their division leads us to

#{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)}# or

#{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)xx(cosbeta-isinbeta)/(cosbeta-isinbeta)}#

#(r_1/r_2){(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/((cos^2beta+sin^2beta))# or

#(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta))# or

#z_1/z_2# is given by #(r_1/r_2, (alpha-beta))#

So for division complex number #z_1# by #z_2# , take new angle as #(alpha-beta)# and modulus the ratio #r_1/r_2# of the modulus of two numbers.

Here #1+7i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt(1^2+7^2)=sqrt50# and #alpha=tan^(-1)7#

and #5-3i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt(5^2+3^2)=sqrt34# and #beta=tan^(-1)(-3/5)#

and #z_1/z_2=sqrt50/(sqrt34)(costheta+isintheta)#, where #theta=alpha-beta#

Hence, #tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=(7-(-3/5))/(1+7xx(-3/5))=(38/5)/(-16/5)=-19/8#.

Hence, #(1+7i)/(5-3i)=sqrt50/sqrt34(costheta+isintheta)#

= #5/sqrt17(costheta+isintheta)#, where #theta=tan^(-1)(-19/8)#

Answer:

The Inverse Sine Function is denoted by, #arc sin # or, #sin^-1# and

is defined by,

#arc sin x=theta, x in [-1,1] iff sin theta=x, theta in [-pi/2,pi/2]#

Explanation:

The Inverse Sine Function is denoted by, #arc sin # or, #sin^-1# and

is defined by,

#arc sin x=theta, x in [-1,1] iff sin theta=x, theta in [-pi/2,pi/2]#

Answer:

#(i-3)/(5i+2)~~0.587*(cos(1.630)+i * sin(-1.630))#

Explanation:

#color(red)("~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~")#
#color(white)("XX ")color(blue)(a+bi harr r * [cos(theta)+i * sin(theta)])#
#color(white)("XXX")color(blue)("where " r=sqrt(a^2+b^2))#
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#

#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

#color(red)("~~~ Trigonometric Division ~~~")#
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]#

#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

If #A=i-3=-3+1i#
then
#color(white)("XXX")r_A=sqrt((-3)^2+1^2)=sqrt(10)#
and
#color(white)("XXX")theta_A=arctan(-1/3)+pi~~2.819842099#

If #B=5i+2=2+5i#
then
#color(white)("XXX")r_B=sqrt(2^2+5^2)=sqrt(29)#
and
#color(white)("XXX")theta_B=arctan(5/2)~~1.19028995#

#A/B=(i-3)/(5+2i)#

#color(white)("XXX")=sqrt(10)/sqrt(29) * ([cos(theta_A)+isin(theta_A))/(cos(theta_B)+isin(theta_B)))#

#color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])#

Plugging in the values for #theta_A# and #theta_B# and using a calculator/spreadsheet:
#color(white)("XXX")~~-0.034482759+0.586206897i#

If we call this value #C#
then
#color(white)("XXX")r_C=sqrt((-0.03448...)^2+(0.5862...)^2)~~0.58722022#
and (since #C# is in Quadrant 2)
#color(white)("XXX")theta_C=arctan((0.5862...)/(-0.03448...))+pi~~1.62955215#

#A/B=C=r_C(cos(theta_C)+i * sin(theta_C))#

Answer:

Use the formula for a circle #(x^2+y^2=r^2)#, and substitute #x=rcostheta# and #y=rsintheta#.

Explanation:

The formula for a circle centred at the origin is

#x^2+y^2=r^2#

That is, the distance from the origin to any point #(x,y)# on the circle is the radius #r# of the circle.

Picture a circle of radius #r# centred at the origin, and pick a point #(x,y)# on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is #r#. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (#theta#).

Now for the trigonometry.

For an angle #theta# in a right triangle, the trig function #sin theta# is the ratio #"opposite side"/"hypotenuse"#. In our case, the length of the side opposite of #theta# is the #y#-coordinate of our point #(x,y)#, and the hypotenuse is our radius #r#. So:

#sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta#

Similarly, #cos theta# is the ratio of the #x#-coordinate in #(x,y)# to the radius #r#:

#cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta#

So we have #x=rcostheta# and #y=rsintheta#. Substituting these into the circle formula gives

#"      "x^2"     "+"      "y^2"     "=r^2#
#(rcostheta)^2+(rsintheta) ^2 = r^2#
#r^2cos^2theta + r^2 sin^2 theta = r^2#

The #r^2#'s all cancel, leaving us with

#cos^2 theta + sin^2 theta = 1#

This is often rewritten with the #sin^2# term in front, like this:

#sin^2 theta + cos^2 theta = 1#

And that's it. That's really all there is to it. Just as the distance between the origin and any point #(x,y)# on a circle must be the circle's radius, the sum of the squared values for #sin theta# and #cos theta# must be 1 for any angle #theta#.

Answer:

#x_1=npi, n=2k+-1, k in ZZ#

#x_2=n/3pi,mod(n,2)=0#

Explanation:

#cos2x+3cosx=-2#

Use the double angle formula for cosine to expand #cos2x# and rewrite the equation in standard form

#2cos^2x-1+3cosx+2=0#

#2cos^2x+3cosx+1=0#

Let #cosx=y#

#y=(-3+-sqrt(9-4(2)(1)))/(2(2))=(-3+-1)/4#

#y=-1# or #y=-1/2#

#cosx=-1#

#x=npi, n=2k+-1, k in ZZ#

#cosx=-1/2#

#x_2=n/3pi,mod(n,2)=0#

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

Taking #(2r-1)pi/8=theta#

# {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

#= {[sintheta]^4 +[costheta]^4}.#

#={[sin^2theta +cos^2theta]^2-2sin^2thetacos^2theta}#

#={1^2-1/2(2sinthetacostheta)^2}#

#={1^2-1/2sin^2 2theta}#

# ={1-1/2sin^2 (2r-1)pi/4}#

# ={1-1/2sin^2 ((rpi)/2-pi/4))}#

# ={1-1/2(pm1/sqrt2)^2}color(red)"*"#

# ={1-1/4 }=3/4#

#[color(red)"*"=>sin((rpi)/2-pi/4)=pmsin(pi/4) or pmcos(pi/4)" for " r in ZZ^"+"]#
So

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

# =sum_(r=1)^(r=4) {3/4}=4xx3/4=3#

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