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Answer:

#theta ~~ 0.76 radians#

Explanation:

Let #barA = <1, 5, 9>#
Let #barB = <-2,3,2>#

The #barA*barB# is:

#barA*barB = (1)(-2) + (5)(3) + (9)(2) = 31#

#|barA| = sqrt(1^2 + 5^2 + 9^2) = sqrt(107)#

#|barB| = sqrt((-2)^2 + 3^2 + 2^2) = sqrt(17)#

Use #barA*barB = |barA||barB|cos(theta)#

Solve for #theta#

#theta = cos^-1((barA*barB)/(|barA||barB|))#

#theta = cos^-1((31)/(sqrt(107)sqrt(17)))#

#theta ~~ 0.76 radians#

Answer:

In polar form #3sqrt2-3sqrt2i# is #6(cos(-pi/4)+isin(-pi/4))#

Explanation:

A complex number #a+bi# can be written in polar form as #r(costheta+isintheta)# i.e. #a=rcostheta# and #b=rsintheta#.

Squaring and adding the last two, we get #a^2+b^2=r^2cos^2theta+r^2sin^2theta#

= #r^2(cos^2theta+sin^2theta)=r^2xx1=r^2#

and #b/a=(rsintheta)/(rcostheta)=tantheta#

Here in the complex number #3sqrt2-3sqrt2i#, we have #a=3sqrt2# and #b=-3sqrt2# and hence #r=sqrt(a^2+b^2)=sqrt((3sqrt2)^2+(-3sqrt2)^2)#

= #sqrt(18+18)=sqrt36=6# and

#costheta=(3sqrt2)/6=1/sqrt2# and #sintheta=(-3sqrt2)/6=-1/sqrt2#

i.e. #theta=(-pi/4)#

Hence in polar form #3sqrt2-3sqrt2i# is #6(cos(-pi/4)+isin(-pi/4))#

Answer:

#x=sqrt(3)/3#

Explanation:

#"arccot"(x) + 2arcsin(sqrt(3)/2) = pi#

The inverse sine function #arcsin(x)# is defined as the unique value in the interval #[-pi/2,pi/2]# such that #sin(arcsin(x)) = x#. On that interval, we have #sin(pi/3) = sqrt(3)/2# as a well known angle. Thus #arcsin(sqrt(3)/2) = pi/3#

#=> "arccot"(x) + (2pi)/3 = pi#

#=> "arccot"(x) = pi/3#

#=> cot("arccot"(x)) = cot((pi)/3)#

#=> x = cot((pi)/3)#

#=cos((pi)/3)/sin((pi)/3)#

#=(1/2)/(sqrt(3)/2)#

#=1/sqrt(3)#

#=sqrt(3)/3#

Answer:

Set your compass to a radius of 6, put the center point at #(6, 0)#, and draw a circle.

Explanation:

Multiply both sides of the equation by r:

#r^2 = 12rcos(theta)#

Substitute #x^2 + y^2# for #r^2# and #x# for #rcos(theta)#

#x^2 + y^2 = 12x#

The standard form of this type of equation (a circle) is:

#(x - h)^2 + (y - k)^2 = r^2#

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract #0# in the square:

#x^2 + (y - 0)^2 = 12x#

To complete the square for the x terms, we add #-12x + h^2# both sides of the equation:

#x^2 - 12x + h^2 + (y - 0)^2 = h^2#

We can use the pattern, #(x - h)^2 = x^2 - 2hx + h^2# to find the value of h:

#x^2 - 2hx + h^2 = x^2 - 12x + h^2#

#-2hx = -12x#

#h = 6#

Substitute 6 for h into the equation of the circle:

#x^2 - 12x + 6^2 + (y - 0)^2 = 6^2#

We know that the first three terms are a perfect square with h = 6:

#(x - 6)^2 + (y - 0)^2 = 6^2#

This is a circle with a radius of 6 and a center point #(6, 0)#

Answer:

#cos(tan^-1(2)+tan^-1(3))=-1/sqrt2#

Explanation:

First use the cosine angle-addition formula:

#cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#

Then the original expression equals:

#=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))#

Note that if #theta=tan^-1(2)#, then #tan(theta)#. That is, a right triangle with angle #theta# has #tan(theta)=2#, which is the triangle with the leg opposite #theta# being #2# and the leg adjacent to #theta# being #1#. The Pythagorean theorem tells us that the hypotenuse of this triangle is #sqrt5#.

So, when #theta=tan^-1(2)#, we see that:

#cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5#

#sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5#

Now let angle #phi# be defined by #phi=tan^-1(3)#, such that #tan(phi)=3#. This is the right triangle with:

#{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}#

Then:

#cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10#

#sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10#

Plugging these into the expression from earlier we get:

#=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)#

Note that #sqrt5(sqrt10)=sqrt50=5sqrt2#:

#=(1-6)/(5sqrt2)=-1/sqrt2#

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