If #sinA + sin^2A=1# and #a cos^12A+ b cos^8A + c cos^6A-1 = 0# then find #(b + c)/(a + b) #?

3 Answers
Dean R.
Jun 2, 2018

That's a scary equation with twelfth powers; let's focus on the other one.

# sin ^2 A + sin A - 1 = 0#

#sin A = 1/2 (-1 pm sqrt{5}) #

Only the positive sign is the sine of a real angle,

#sin A = 1/2(sqrt{5}-1)#

#sin ^2 A = 1/4( 6 - 2 sqrt{5}) = 1/2(3- sqrt{5}) #

#cos ^2 A = 1 - sin ^2 A = 1/2 (sqrt{5}-1) = sin A #

#cos ^4 A = sin ^2 A = 1/2(3 - sqrt{5}) #

#cos ^6 A = (1/2 (sqrt{5}-1))( 1/2(3 - sqrt{5}) ) = sqrt{5} - 2 #

#cos ^8 A = (sqrt{5}-1)( 1/2 (sqrt{5}-1) ) =3 - sqrt{5} #

#cos^12 A = (cos ^6 A)^2 = (sqrt{5} - 2)^2=9-4sqrt{5} #

Now the equation becomes

# a ( 9-4sqrt{5} ) + b(3 - sqrt{5}) + c(sqrt{5} - 2 ) -1 = 0#

That's a plane in #a,b,c# space.

# c = (2 + sqrt(5)) ((4 sqrt(5) - 9) a + (sqrt(5) - 3) b + 1)#

#(a + b)/( b+c) #

# = (a + b)/( b + ( (2 + sqrt(5)) ((4 sqrt(5) - 9) a + (sqrt(5) - 3) b + 1) ) ) #

#= (a + b)/( (2-sqrt(5)) a - b sqrt(5) + sqrt(5) + 2)#

Not that satisfying.

P dilip_k
Jun 2, 2018

Given

#sinA+sin^2A=1#

#=>sinA=1-sin^2A#

#=>sin^2A=(cos^2A)^2#

#=>1-cos^2A=cos^4A#

#=>1-cos^4A=cos^2Acolor(red)(....[1]#

Now

#=>(cos^4A+cos^2A)^3=1^3#

#=>cos^12A+3*(cos^4A)^2*cos^2A+3*cos^4A*(cos^2A)^2+(cos^2A)^3=1#

#=>cos^12A+3cos^8A(cos^2A+1)+cos^6A-1=0#

#=>cos^12A+3cos^8A(1-cos^4A+1)+cos^6A-1=0#
#color(red)(["inserting " cos^2A=1-cos^4A" from "[1]])#

#=>cos^12A+3cos^8A(2-cos^4A)+cos^6A-1=0#

#=>cos^12A+6cos^8A-3cos^12A+cos^6A-1=0#

#=>-2cos^12A+6cos^8A+cos^6A-1=0color(red)(...[2])#

Given relation

#acos^12A+bcos^8A+c cos^6A-1=0#

Comparing this relation with [2] we get

#a=-2,b=6andc=1#

Hence the required value of the given expression

#(b+c)/(a+b)#

#=(6+1)/(-2+6)=7/4#

George C.
Jun 3, 2018

If #a, b, c# are integers, then for some integer #k#, we have:

#(a, b, c) = (1 - k, 2k, 4-k)#

and:

#(b + c)/(a + b) = (k+4)/(k+1)#

Explanation:

Given:

#sinA + sin^2A=1#

We have:

#0 = 4(sin^2A + sin A-1)#

#color(white)(0) = 4sin^2A+4sinA+1-5#

#color(white)(0) = (2sinA+1)^2-(sqrt(5))^2#

#color(white)(0) = (2sinA+1-sqrt(5))(2sinA+1+sqrt(5))#

So:

#sinA = (-1+sqrt(5))/2" "# or #" "sinA = (-1-sqrt(5))/2#

We can discount the latter since it gives a value out of the range #[-1, 1]#, so is not satisfied by any Real value of #A#.

So:

#sinA = -1/2+sqrt(5)/2#

Then:

#sin^2A = (-1/2+sqrt(5)/2)^2 = 1/4-sqrt(5)/2+5/4 = 3/2-sqrt(5)/2#

So:

#cos^2A = 1-sin^2A = 1-(3/2-sqrt(5)/2) = -1/2+sqrt(5)/2#

Also:

#cos^4A = (cos^2A)^2 = (-1/2+sqrt(5)/2)^2 = 3/2-sqrt(5)/2#

#cos^6A = (cos^2A)(cos^4A)#

#color(white)(cos^6A) = (-1/2+sqrt(5)/2)(3/2-sqrt(5)/2)#

#color(white)(cos^6A) = -3/4+sqrt(5)/4+3sqrt(5)/4-5/4#

#color(white)(cos^6A) = -2+sqrt(5)#

#cos^8A = (cos^4 A)^2 = (3/2-sqrt(5)/2)^2 = 9/4-(3sqrt(5))/2+5/4 = 7/2-(3sqrt(5))/2#

#cos^12A = (cos^6A)^2 = (-2+sqrt(5))^2 = 4-4sqrt(5)+5 = 9-4sqrt(5)#

Then:

#0 = 2(a cos^12A+ b cos^8A + c cos^6A-1)#

#color(white)(0) = 2(a(9-4sqrt(5))+b(7/2-3/2sqrt(5))+c(-2+sqrt(5))-1)#

#color(white)(0) = (18a+7b-4c-2)+(-8a-3b+2c)sqrt(5)#

This defines a plane of irrational slope in #a, b, c# coordinate space.

It is insufficient to determine a unique value for #(b + c)/(a + b) #

Perhaps the question is missing some specification, e.g. that #a, b, c# are integers.

What integer lattice points lie on the plane?

They must satisfy:

#{ (18a+7b-4c-2=0), (-8a-3b+2c=0) :}#

Adding twice the second equation to the first, we find:

#2a+b-2=0#

So in order that #a# be an integer, we need #b# to be even, i.e. #b=2k# for some integer #k#

Then:

#a=1/2(2-b) = 1/2(2-2k) = 1-k#

and we find:

#c = 1/2(8a+3b) = 1/2(8-8k+6k) = 4-k#

Then:

#(b + c)/(a + b) = (2k+(4-k))/((1-k)+2k) = (k+4)/(k+1)#

For example, with #k=3# we find:

#a=-2#, #b=6#, #c=1# and #(b+c)/(a+b) = (k+4)/(k+1) = 7/4#

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