Featured Answers

8
Active contributors today

Answer:

Polar coordinate #(2,(5pi)/6)# is a distance of #2# from the origin
and at an angle of #(5pi)/6#

Explanation:

Note that the angle #(5pi)/6# has a reference angle of #pi/6# above the #pi# axis.
#pi/6# is a common angle with hypotenuse #=2# and opposite side #=1#
enter image source here

Answer:

Please see the explanation.

Explanation:

Here is the graph:

Desmos.com

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

#f(x) = Acos(x - phi) = sin(x) + cos(x)#

My graphing tool allows me to obtain values of points and I can tell you that #A = sqrt(2)#

#cos(x - phi) = 1/sqrt(2)sin(x) + 1/sqrt(2)cos(x)#

This fits the trigonometric identity:

#cos(x - phi) = sin(x)sin(phi) + cos(x)cos(phi)#

where #cos(phi) = sin(phi) = 1/sqrt(2)#

This happens at #phi = pi/4#

The graphing tool confirms that the x coordinates are shift by #pi/4#.

#f(x) = sqrt(2)cos(x - pi/4)#

Substitute #f^-1(x)# for every x:

#f(f^-1(x)) = sqrt(2)cos(f^-1(x) - pi/4)#

The left side becomes x by definition:

#x = sqrt(2)cos(f^-1(x) - pi/4)#

Make the #sqrt(2)# on the right disappear by multiplying both sides by #sqrt(2)/2#:

#sqrt(2)/2x = cos(f^-1(x) - pi/4)#

Use the inverse cosine on both sides:

#cos^-1(sqrt(2)/2x) = f^-1(x) - pi/4#

Solve for #f^-1(x)#:

#f^-1(x) = cos^-1(sqrt(2)/2x) + pi/4#

To confirm that this is truly an inverse, verify that #f(f^-1(x)) = f^-1(f(x)) = x#

Answer:

See below:

Explanation:

With a triangle with #sectheta=2#, let's first remember that:

#sectheta="hypotenuse"/"adjacent"=2/1#

We could run through the pythagorean theorem for the third side, but we can also remember that these two measurements are part of a 30-60-90 triangle, and so the third side is #sqrt3#. To prove it, let's go ahead and do Pythagorean Theorem:

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

#1^2+(sqrt3)^2=2^2#

#1+3=4#

This gives us the 6 trig ratios:

#sintheta=sqrt3/2#

#costheta=1/2#

#tantheta=(sqrt3/2)/(1/2)=(sqrt3/2)(2/1)=sqrt3#

#csctheta=2/sqrt3=(2sqrt3)/3#

#sectheta=2/1=2#

#cottheta=1/sqrt3=sqrt3/3#

The angle of the triangle we're looking at is #60=pi/3#, with the opposite being the middle length of #sqrt3#, the adjacent length of 1, and hypotenuse of 2.

freemathhelp.com

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:

#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#

Answer:

Here is the graph of #y=sin^(-1)x#.

enter image source here

Explanation:

Step 1: Sketch the graph of #y=sin x# on #[-pi/2,pi/2]#.

enter image source here

Step 2: Sketch the line #y=x#.

enter image source here

Step 3: Reflect the graph of #y=sin x# about the line #y=x#.

enter image source here

View more
Questions
Ask a question Filters
Loading...
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions