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Featured 3 months ago

Let

Let

The

Use

Solve for

Featured 3 months ago

In polar form

A complex number

Squaring and adding the last two, we get

=

and

Here in the complex number

=

i.e.

Hence in polar form

Featured 3 months ago

The inverse sine function

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

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

#=1/sqrt(3)#

#=sqrt(3)/3#

Featured 3 months ago

Set your compass to a radius of 6, put the center point at

Multiply both sides of the equation by r:

Substitute

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

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

To complete the square for the x terms, we add

We can use the pattern,

Substitute 6 for h into the equation of the circle:

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

This is a circle with a radius of 6 and a center point

Featured 2 weeks ago

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

So, when

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

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

Now let angle

#{("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

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

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