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Answer:

#sinu#.

Explanation:

Letting, #(u-v)=w,# and, knowing that,

#sin(x+y)=sinxcosy+cosxsiny,# we have,

The Expression #=sinwcosv+coswsinv#

#=sin(w+v)#

#=sin((u-v)+v).................[because, w=u-v]#

#=sinu#

Answer:

The distance is #~~2.7#

Explanation:

Change the -1 radius to +1 by adding #pi# to the angle:

#(1, (4pi)/3)#

Find the difference between the two angles:

#theta = (4pi)/3 - (5pi)/8#

#theta = (32pi)/24 - (15pi)/24#

#theta = (17pi)/24#

We can find the distance between them using the Law of Cosines where side #a = 2#, side #b =1# and the angle between them is #theta = (17pi)/24#

#c^2 = 2^2 + 1^2 - 2(2)(1)cos((17pi)/24)#

#c ~~ 2.7#

Let #a+b=x and a-b =y#

So #2a=x+y and 2b =x-y#

Now #RHS=cos2b-cos2a#

#=cos(x-y)-cos(x+y)#

#=cosxcosy+sinxsiny-cosxcosy+sinxsiny#

#=2sinxsiny#

#=2sin(a+b)sin(a-b)=LHS#

Answer:

THe angle is 66.2º

Explanation:

The dot product betwwen 2 vectors is
#〈veca〉.〈vecc〉=∥〈veca〉∥∥〈vecc〉∥cos(veca,vecc)#

#cos(veca,vecc)=(〈veca〉 .〈 vecc 〉 ) /(∥〈veca〉∥∥〈vecc〉∥)#

If #〈veca〉=〈a_1,a_2,a_3〉#
and #〈vecc〉=〈c_1,c_2,c_3〉#

The dot product is #〈veca〉.〈vecc〉=a_1c_1+a_2c_2+a_3c_3#

#〈vecc〉=〈veca〉-〈vecb〉 =〈2,1,8〉-〈5,7,3〉 =〈-3,-6,5〉#

#〈veca〉.〈vecc〉=-6+(-6)+40=28#

#∥〈veca〉∥=sqrt(a_1^2+a_2^2+a_3^2)#

#∥〈veca〉∥=sqrt(4+1+64)=sqrt69#

#∥〈vecc〉∥=sqrt(9+36+25)=sqrt70#
let #cos(veca,vecc)=costheta#
so the angle
#costheta=28/((sqrt69)(sqrt70))=0.403#
#theta =66.2º#

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.

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Answer:

The explanation is written with the assumption that you meant #theta# for the argument of the cosine function.

Explanation:

Use the distributive property :

#2r - rcos(theta) = 2#

#2r = rcos(theta) + 2#

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

#2sqrt(x^2 + y^2) = x + 2#

Square both sides:

#4(x^2 + y^2) = x^2 + 4x + 4#

#4x^2 + 4y^2 = x^2 + 4x + 4#

#3x^2 - 4x + 4y^2 = 4#

Add #3h^2# to both sides:

#3x^2 - 4x + 3h^2 + 4y^2 = 3h^2+ 4#

Remove a factor of the 3 from the first 3 terms:

#3(x^2 - 4/3x + h^2) + 4y^2 = 3h^2+ 4#

Use the middle in the right side of the pattern #(x - h)^2 = x^2 - 2hx + h^2# and middle term of the equation to find the value of h:

#-2hx = -4/3x#

#h = 2/3#

Substitute the left side of the pattern into the equation:

#3(x - h)^2 + 4y^2 = 3h^2+ 4#

Substitute #2/3# for h and insert a -0 into the y term:

#3(x- 2/3)^2 + 4(y-0)^2 = 3(2/3)^2+ 4#

#3(x- 2/3)^2 + 4(y-0)^2 = 16/3#

Divide both sides by #16/3#

#3(x- 2/3)^2/(16/3) + 4(y-0)^2/(16/3) = 1#

#(x- 2/3)^2/(16/9) + (y-0)^2/(16/12) = 1#

Write the denominators as squares:

#(x- 2/3)^2/(4/3)^2 + (y-0)^2/(4/sqrt(12))^2 = 1#

The is standard Cartesian form of the equation of an ellipse with a center at #(2/3,0)#; its semi-major axis is #4/3# units long and is parallel to the x axis and its semi-minor axis is #4/sqrt(12)# units long and is parallel to the y axis.

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