Question
If #sinx+sin^2x=1# and #acos^12x+bcos^8x+c cos^6x-1=0#, then what is the value of #a^2+b^2+c^2#?
Given
#sinx+sin^2x=1#
#=>sin^2x+sinx-1=0#
#=>sinx=(-1pmsqrt(1^2-4*1*(-1)))/2#
#=>sinx=(-1pmsqrt5)/2#
But we know #-1<=sinx<=1#
Hence acceptable value of
#sinx=(-1+sqrt5)/2=1/2(sqrt5-1)#
Now
#color(red)(sinx=1/2(sqrt5-1))#
#color(red)(sin^2x=1/4(5+1-2sqrt5)=1/2(3-sqrt5)#
#color(blue)(sin^3x=sinx xx sin^2x=1/2(sqrt5-1)xx1/2(3-sqrt5))#
#=>color(blue)(sin^3x=(-2+sqrt5))#
#color(magenta)(sin^4x=(1/2(3-sqrt5))^2=(7/2-3/2sqrt5)#
#color(green)(sin^6x=(-2+sqrt5)^2=(9-4sqrt5))#
Again we have
#sinx+sin^2x=1#
#=>sinx=1-sin^2x=cos^2x#
Now putting #cos^2x=sinx # in the LHS of the given expression
#asin^6x+bsin^4x+c sin^3x-1=0#
Now putting the values of #sin^6x " "sin^4x and sin^3x#
#a(9-4sqrt5)+b(7/2-3/2sqrt5)+c (-2+sqrt5)-1=0#
Now if we take rationalizing factors
#a=(9+4sqrt5)#
#b=(7/2+3/2sqrt5)#
and
#c= (-2-sqrt5)#
then above equation is satisfied.
So inserting these values we can get a possible value of
#a^2+b^2+c^2#
#=(9+4sqrt5)^2+(7/2+3/2sqrt5)^2+ (-2-sqrt5)^2#
#=(161+72sqrt5)+1/4(94+42sqrt5)+ (9+4sqrt5)#
#=(161+72sqrt5)+(23.5+10.5sqrt5)+ (9+4sqrt5)#
#=(193.5+86.5sqrt5)#
In addition here is a scratchpad note of respected Cesareo R regarding this problem .
https://socratic.org/scratchpad/87db6c91b6866e1f5e3a
