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#?

1 Answer
Apr 8, 2017

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