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Answer:

(see below)

Explanation:

Based on the Pythagorean Theorem, we know that
#color(white)("XXX")a^2+b^2=c^2

#rarr b=sqrt(c^2-a^2)=sqrt(6^2-5^2)=sqrt(11)#
enter image source here

#sin = "opposite"/"hypotenuse"color(white)("XX")rarr sin(A)=5/6#

#csc = "hypotenuse"/"opposite" color(white)("XX")rarr csc(A)=6/5#

#cos = "adjacent"/"hypotenuse"color(white)("XX")rarr cos(A)=sqrt(11)/6#

#sec = "hypotenuse"/"adjacent" color(white)("XX")rarr sec(A)=6/sqrt(11)#

#tan = "opposite"/"adjacent"color(white)("XX")rarr tan(A)=5/sqrt(11)#

#cot = "adjacent"/"opposite" color(white)("XX")rarr cot(A)=sqrt(11)/5#

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)#

Given expression #=arcsin((x+1)/sqrt(2*(x²+1)))#
Let #x = tantheta#

So #theta=tan^-1x#

Inserting #x = tantheta# the given expression becomes

#=arcsin((tantheta+1)/sqrt(2*(tan^2theta+1)))#

#=arcsin((sintheta/costheta+1)/sqrt(2*(sec^2theta)))#

#=arcsin(1/sqrt2((sinthetasectheta+1)/(sectheta)))#

#=arcsin(1/sqrt2((sinthetacancelsectheta)/cancelsectheta+1/(sectheta)))#

#=arcsin(1/sqrt2(sintheta+costheta))#

#=arcsin(1/sqrt2sintheta+1/sqrt2costheta)#

#=arcsin(cos(pi/4)sintheta+sin(pi/4)costheta)#

#=arcsin(sin(theta+pi/4))#

#=theta+pi/4#

#=tan^-1x+pi/4#

Answer:

#x in [pi/6, (5pi)/6]#

Explanation:

Note first that as #-1 <= sinx <= 1# for all real #x#, we must have #sinx-2 <= -1 < 0#. Then, #(sinx-1/2)(sinx-2)# is the product of #sinx-1/2# and a negative number, meaning it is less than or equal to #0# if and only if #sinx-1/2 >= 0#.

Thus, the given problem is equivalent to the problem #sinx-1/2 >= 0# with the restriction #x in [0, 2pi]#

Adding #1/2# to each side of the inequality, we get

#sinx >= 1/2#

Using the unit circle, we can tell that on our restricted interval, we have #sinx>=1/2# if and only if #x in [pi/6, (5pi)/6]#. Therefore, this is our answer.

#x in [pi/6, (5pi)/6]#

You have to start by finding a point that lies on this line. I think the simplest would be #(-6, -1)# (since we are in quadrant III, both the x and y axis have to have negative values).

We now draw an imaginary triangle, as shown in the following diagram.

enter image source here

We can now clearly see that our side opposite #theta# measures #1# unit and our side adjacent #theta# measures #6# units.

We must finally find the hypotenuse prior to determining the ratios. By pythagorean theorem:

#a^2 + b^2 = c^2#

#(-1)^2 + (-6)^2= c^2#

#1 + 36 = c^2#

#c^2 = 37#

#c = +-sqrt(37)#

However, the hypotenuse can never have a negative length, so we can only accept the positive solution. We can now find our ratios.

#sintheta = "opposite"/"hypotenuse" = -1/sqrt(37) = -sqrt(37)/37#

#csctheta = 1/sintheta = 1/("opposite"/"hypotenuse") = "hypotenuse"/"opposite" = sqrt(37)/(-1) = -sqrt(37)#

#costheta = "adjacent"/"hypotenuse" = -6/sqrt(37) = (-6sqrt(37))/37#

#sectheta = 1/costheta = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent" = sqrt(37)/(-6) = -sqrt(37)/6#

#tantheta = "opposite"/"adjacent" = -1/(-6) = 1/6#

#cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = -6/(-1) = 6#

Hopefully this helps!

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))#

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